Quadratic Equations Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 4.1
  1. x + 1/x = 2 , x not equal 0 Examine whether the following equations are…
  2. (x — 2)(x + 3) = 0 Examine whether the following equations are quadratic…
  3. 2x^2 - root 5x+2 = 0 Examine whether the following equations are quadratic…
  4. 1/x+1 - 1/x-1 = 3 (x not equal plus or minus 1) Examine whether the following…
  5. (2x + 1)(2x — 1) = (4x + 3)(x — 5) Examine whether the following equations are…
  6. x-1/x+1 - x+1/x-1 = 2/2 (x not equal plus or minus 1) Examine whether the…
  7. (2x + 3)^2 — (3x + 2)^2 = 13 Examine whether the following equations are…
  8. x^2 — 3x + 2 = 0, x = 2 Verify whether the given value of x is a solution of…
  9. x^2 + x — 2 = 0, x = 2 Verify whether the given value of x is a solution of the…
  10. 1/3x+1 - 1/2x-1 + 3/4 = 0 , x = 1[x not equal 1/2 , - 1/3] Verify whether the…
  11. (3x — 8)(2x + 5) = 0, x = - Verify whether the given value of x is a solution…
  12. If x = 1 is a root of ax^2 + bx + c = 0, a not equal 0, a, b, c e R, prove that…
  13. If x = —1 is a root of x^2 — px + q = 0, p, q ∈ R, prove that p + q + 1 = O.…
  14. Find k, if one of the roots of x^2 — kx + 6 = 0 is 3.
  15. Find k, if one of the roots of x^2 + 3(k + 2)x — 9 = 0 is —3.
  16. 27x^2 — 48 = 0 Solve the following equations using the method of factorization:…
  17. (x 7)^2 16 = 0 Solve the following equations using the method of factorization:…
  18. 6x^2 + 13x + 6 = 0 Solve the following equations using the method of…
  19. 15x^2 16x + 1 = 0 Solve the following equations using the method of…
  20. root 5x^2 - 4x - root 5 = 0 Solve the following equations using the method of…
  21. x + 1/x = 2 1/6 Solve the following equations using the method of…
Exercise 4.2
  1. 6x^2 — 13x + 6 = 0 Find the discriminant of the following quadratic equations…
  2. √6 x^2 — 5x + √6 = 0 Find the discriminant of the following quadratic equations…
  3. 24x^2 — 17x + 3 = 0 Find the discriminant of the following quadratic equations…
  4. x^2 + 2x + 4 = 0 Find the discriminant of the following quadratic equations and…
  5. x^2 + x + 1 = 0 Find the discriminant of the following quadratic equations and…
  6. x^2 — 3√3x — 30 = 0 Find the discriminant of the following quadratic equations…
  7. If a, b, c R, a 0, c 0, then prove that the roots of ax^2 + bx + c = 0 are real…
  8. Find k, if the roots of x^2 — (3k — 2)x + 2k = 0 are equal and real.…
  9. If the roots of the quadratic equation (k + 1)x^2 — 2(k—1)x + 1 = 0 are real…
  10. If the roots of ax^2 + 2bx + c = 0, a ≠ 0, a, b, c ∈ R are real and equal, then…
  11. x^2 + 10x + 6 = 0 Solve the following equations using the general formula:…
  12. x^2 + 5x — 1 = 0 Solve the following equations using the general formula:…
  13. x^2 — 3x — 2 = 0 Solve the following equations using the general formula:…
  14. x^2 — 3√6x + 12 = 0 Solve the following equations using the general formula:…
  15. 3x^2 + 5√2x + 2 = 0 Solve the following equations using the general formula:…
  16. x^2 - 1/x^2 + 1 = 4/5 Solve the following equations using the general formula:…
Exercise 4.3
  1. Find two numbers whose sum is 27 and the product is 182.
  2. Find two consecutive natural numbers, sum of whose squares is 365.…
  3. The sum of ages of two friends is 20 years. Four years ago the product of their…
  4. A rectangular garden is designed such that the length of the garden is twice its…
  5. Perimeter of a rectangular garden is 360 m and its area is 8000 m^2 . Find the…
  6. If a cyclist travels at a speed 2 km/hr more than his usual speed, he reaches…
  7. The diagonal of a rectangular ground is 60 meters more than the breadth of the…
  8. The sides of a right angled triangle are consecutive positive integers. Find the…
Exercise 4
  1. x^2 — 12 = 0 Solve the following quadratic equations using factorization:…
  2. x^2 — 7x — 60 = 0 Solve the following quadratic equations using factorization:…
  3. x^2 — 15x + 56 = 0 Solve the following quadratic equations using factorization:…
  4. 2x+3/2x-3 + 2x-3/2x+3 = 17/4 , x not equal 3/2 Solve the following quadratic…
  5. 1/x+5 + 3/4 (3x+1) = 1/x+2 , x not equal -5 x not equal -2 , x not equal - 1/3…
  6. x^2 — 24x — 16 = 0 Find the roots of the following equations by the method of…
  7. 3x^2 + 7x — 20 = 0 Find the roots of the following equations by the method of…
  8. x^2 - 10x + 25 = 0 Find the roots of the following equations by the method of…
  9. x^2 + (x + 5)^2 = 625 Find the roots of the following equations by the method…
  10. (x + 2)(x + 3) = 240 Find the roots of the following equations by the method of…
  11. Divide 20 into two parts such that the sum of the square of the parts is 218.…
  12. A car takes 1 hour less to cover a distance of 200 km if its speed is increased…
  13. When there is a decrease of 5 km/hr in the usual uniform speed of a goods train,…
  14. A river flows at a speed of 1 km/hr. A boat takes 15 hours to travel 112 km…
  15. Find a number greater than 1 such that the sum of the number and its reciprocal…
  16. The difference of the speed of a faster car and a slower car is 20 km/hr. If the…
  17. Product of the ages of Virat 7 years ago and 7 years later is 480. Find his…
  18. If the age of Sachin 8 year ago is multiplied by his age two years later, the…
  19. Sunita's age at present is 2 years less than 6 times the age of her daughter…
  20. The formula of the sum of first n natural numbers is S = n (n+1)/2 . If the sum…
  21. Hypotenuse of a right angled triangle is 2 less than 3 times its shortest side.…
  22. The sum of the squares of two consecutive odd positive integers is 290. Find…
  23. The product of two consecutive even natural numbers is 224. Find the numbers.…
  24. The product of digits of a two - digit number is 8 and the sum of the squares…
  25. If price of sugar decreases by 5, one can buy 1 kg more sugar in ! 150, what is…
  26. If the price of petrol is increased by Rs. 5 per litre. One gets 2 litres less…
  27. A vendor gets a profit in percentage equal to the cost price of a flower pot…
  28. While selling a pen for Rs. 24 the loss in percentage is equal to its cost…
  29. The difference of lengths of sides forming right angle in right angled triangle…
  30. The sides of a right angled triangle are x, x + 3, x + 6, x being a positive…
  31. …………… is a solution of quadratic equation x^2 — 3x + 2 = 0A. —3 B. 1 C. 3 D.…
  32. Discriminant D = ….for the quadratic equation 5x^2 — 6x + 1 = 0A. 16 B. root…
  33. If x = 2 is a root of the equation x^2 — 4x + a = 0, then a =A. —2 B. 2 C. —4…
  34. A quadratic equation has two equal roots, if…….A. D 0 B. D 0 C. D = 0 D. D is…
  35. The quadratic equation has 3 as one of its roots.A. x^2 — x — 6 = 0 B. x^2 + x…
  36. If 4 is a root of quadratic equation x^2 + ax — 8 = 0, then a =A. 2 B. 4 C. —2…
  37. If one of the roots of kx^2 — 7x + 3 = 0 is 3, then k =A. —2 B. 3 C. —3 D. 2…
  38. The discriminate of x^2 — 3x — k = 0 is 1. A value of x isA. —4 B. —2 C. 2 D.…

Exercise 4.1
Question 1.

Examine whether the following equations are quadratic equations or not:



Answer:

x +  = 2


⇒  = 2


⇒ x2 + 1 = 2x


⇒ x2 – 2x + 1 = 0


Comparing equation x2 – 2x + 1 = 0 with ax2 + bx + c = 0 we get


a = 1, b = – 2 and c = 1


Here a, b, c ∈ R and a ≠ 0


The degree of polynomial x2 – 2x + 1 = 0 is 2


Hence x +  = 2 is a quadratic equation



Question 2.

Examine whether the following equations are quadratic equations or not:

(x — 2)(x + 3) = 0


Answer:

⇒ x2 + 3x – 2x – 6 = 0


⇒ x2 + x – 6 = 0


Comparing equation x2 + x – 6 = 0 with ax2 + bx + c = 0 we get


a = 1, b = 1 and c = – 6


Here a, b, c ∈ R and a ≠ 0


The degree of polynomial x2 + x – 6 = 0 is 2


Hence (x — 2)(x + 3) = 0 is a quadratic equation



Question 3.

Examine whether the following equations are quadratic equations or not:



Answer:

2x2 – √5x + 2 = 0


Comparing equation 2x2 – √5x + 2 = 0 with ax2 + bx + c = 0 we get


a = 2, b = – √5 and c = 2


Here a, b, c ∈ R and a ≠ 0


The degree of polynomial 2x2 – √5x + 2 = 0 is 2


Hence 2x2 – √5x + 2 = 0 is a quadratic equation



Question 4.

Examine whether the following equations are quadratic equations or not:



Answer:




For simplification of denominator use formula (a + b)(a – b) = a2 – b2



⇒ – 2 = 3 × (x2 – 1)


⇒ – 2 = 3x2 – 3


⇒ 3x2 – 3 + 2 = 0


⇒ 3x2 + 0x – 1 = 0


Comparing equation 3x2 + 0x – 1 = 0 with ax2 + bx + c = 0 we get


a = 3, b = 0 and c = – 1


Here a, b, c ∈ R and a ≠ 0


The degree of polynomial 3x2 + 0x – 1 = 0 is 2


Hence  is a quadratic equation



Question 5.

Examine whether the following equations are quadratic equations or not:

(2x + 1)(2x — 1) = (4x + 3)(x — 5)


Answer:

For L.H.S use formula (a + b)(a – b) = a2 – b2


⇒ 4x2 – 1 = 4x2 – 20x + 3x – 15


⇒– 1 = – 17x – 15


⇒ – 1 + 17x + 15 = 0


⇒ 17x + 14 = 0


⇒ 0x2 + 17x + 14 = 0


Comparing equation 0x2 + 17x + 14 = 0 with ax2 + bx + c = 0 we get


a = 0, b = 17 and c = 14


Here a, b, c ∈ R but a = 0


Therefore, the degree of polynomial 0x2 + 17x + 14 = 0 becomes 1


Hence (2x + 1)(2x — 1) = (4x + 3)(x — 5) is not a quadratic equation



Question 6.

Examine whether the following equations are quadratic equations or not:



Answer:




Expand numerator using formula (a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2 and denominator using formula (a + b)(a – b) = a2 – b2






⇒ – 4x = x2 – 1


⇒ x2 + 4x – 1 = 0


Comparing equation x2 + 4x – 1 = 0 with ax2 + bx + c = 0 we get


a = 1, b = 4 and c = – 1


Here a, b, c ∈ R and a ≠ 0


The degree of polynomial x2 + 4x – 1 = 0 is 2


Hence  –  =  is a quadratic equation



Question 7.

Examine whether the following equations are quadratic equations or not:

(2x + 3)2 — (3x + 2)2 = 13


Answer:

Using the identity (a + b)(a – b) = a2 – b2


Where a = (2x + 3) and b = (3x + 2)


⇒ [(2x + 3) + (3x + 2)] × [(2x + 3) – (3x + 2)] = 13


⇒ (5x + 5) × [2x + 3 – 3x – 2] = 13


⇒ (5x + 5) × (1 – x) = 13


⇒ 5x – 5x2 + 5 – 5x = 13


⇒ – 5x2 + 5 – 13 = 0


⇒ – 5x2 + 0x – 8 = 0


Comparing equation – 5x2 + 0x – 8 = 0 with ax2 + bx + c = 0 we get


a = – 5, b = 0 and c = – 8


Here a, b, c ∈ R and a ≠ 0


The degree of polynomial – 5x2 + 0x – 8 = 0 is 2


Hence (2x + 3)2 — (3x + 2)2 = 13 is a quadratic equation



Question 8.

Verify whether the given value of x is a solution of the quadratic equation or not:

x2 — 3x + 2 = 0, x = 2


Answer:

x = 2 will be solution of quadratic equation x2 — 3x + 2 = 0 if x = 2 satisfies the given equation


put x = 2 in L.H.S of x2 — 3x + 2 = 0


⇒ L.H.S = 22 – 3 × 2 + 2


⇒ L.H.S = 4 – 6 + 2


⇒ L.H.S = – 2 + 2


⇒ L.H.S = 0


⇒ L.H.S = R.H.S


Thus x = 2 satisfies the equation x2 — 3x + 2 = 0 hence x = 2 is a solution



Question 9.

Verify whether the given value of x is a solution of the quadratic equation or not:

x2 + x — 2 = 0, x = 2


Answer:

x = 2 will be solution of quadratic equation x2 + x — 2 = 0 if x = 2 satisfies the given equation


put x = 2 in L.H.S of x2 + x — 2 = 0


⇒ L.H.S = 22 – 2 + 2


⇒ L.H.S = 4


⇒ L.H.S ≠ R.H.S


Thus x = 2 does not satisfies the equation x2 + x — 2 = 0 hence x = 2 is not a solution



Question 10.

Verify whether the given value of x is a solution of the quadratic equation or not:



Answer:


x = 1 will be solution of quadratic equation 


if x = 1 satisfies the given equation


put x = 1 in L.H.S of 1/(3x + 1) – 1/(2x – 1) + 3/4 = 0


⇒ L.H.S 


⇒ L.H.S 


⇒ L.H.S 


⇒ L.H.S 


⇒ L.H.S = 0


⇒ L.H.S = R.H.S


Thus x = 1 satisfies the equation  hence x = 1 is a solution



Question 11.

Verify whether the given value of x is a solution of the quadratic equation or not:

(3x — 8)(2x + 5) = 0, x = – 


Answer:

 will be solution of quadratic equation (3x — 8)(2x + 5) = 0 if satisfies the given equation


put  in L.H.S of (3x — 8)(2x + 5) = 0


⇒ L.H.S 


⇒ L.H.S ]


⇒ L.H.S 


⇒ L.H.S = 0


⇒ L.H.S = R.H.S


Thus  satisfies the equation (3x — 8)(2x + 5) = 0 hence is a solution



Question 12.

If x = 1 is a root of ax2 + bx + c = 0, a  0, a, b, c R, prove that a + b + c = O.


Answer:

As x = 1 is a root of ax2 + bx + c = 0 hence x = 1 will satisfy the equation ax2 + bx + c = 0


Put x = 1 in ax2 + bx + c = 0


⇒ a × (1)2 + b × (1) + c = 0


⇒ a + b + c = 0


Hence proved



Question 13.

If x = —1 is a root of x2 — px + q = 0, p, q ∈ R, prove that p + q + 1 = O.


Answer:

As x = – 1 is a root of x2 — px + q = 0 hence x = – 1 will satisfy the equation x2 — px + q = 0


Put x = – 1 in x2 — px + q = 0


⇒ (– 1)2 – p × (– 1) + q = 0


⇒ 1 + p + q = 0


⇒ p + q + 1 = 0


Hence proved



Question 14.

Find k, if one of the roots of x2 — kx + 6 = 0 is 3.


Answer:

One of the root of equation x2 — kx + 6 = 0 is 3 which means x = 3 is a root


As x = 3 is a root of x2 — kx + 6 = 0 hence x = 3 will satisfy the equation x2 — kx + 6 = 0


Put x = 3 in equation x2 — kx + 6 = 0


⇒ 32 – k × (3) + 6 = 0


⇒ 9 – 3k + 6 = 0


⇒ 15 – 3k = 0


⇒ 15 = 3k


⇒ k = 


⇒ k = 3



Question 15.

Find k, if one of the roots of x2 + 3(k + 2)x — 9 = 0 is —3.


Answer:

One of the root of equation x2 + 3(k + 2)x — 9 = 0 is – 3 which means x = – 3 is a root


As x = – 3 is a root of x2 + 3(k + 2)x — 9 = 0 hence x = – 3 will satisfy the equation x2 + 3(k + 2)x — 9 = 0


Put x = – 3 in equation x2 + 3(k + 2)x — 9 = 0


⇒ (– 3)2 + [3 × (k + 2) × (– 3)] – 9 = 0


⇒ 9 + (– 9) × (k + 2) – 9 = 0


⇒ (– 9) × (k + 2) = 0


⇒ k + 2 = 0


⇒ k = – 2



Question 16.

Solve the following equations using the method of factorization:

27x2 — 48 = 0


Answer:

27 and 48 are divisible by 3 take factor 3 common


⇒ 3 × (9x2 – 16) = 0


⇒ 9x2 – 16 = 0


⇒ (3x)2 – 42 = 0


Using the identity (a + b)(a – b) = a2 – b2


Where a = 3x and b = 4


⇒ (3x + 4)(3x – 4) = 0


⇒ 3x + 4 = 0 or 3x – 4 = 0


⇒ 3x = – 4 or 3x = 4


Therefore  and 



Question 17.

Solve the following equations using the method of factorization:

(x — 7)2 — 16 = 0


Answer:

(x — 7)2 — 42 = 0


Using the identity (a + b)(a – b) = a2 – b2


Where a = (x – 7) and b = 4


⇒ (x – 7 + 4)(x – 7 – 4) = 0


⇒ (x – 3)(x – 11) = 0


⇒ x – 3 = 0 or x – 11 = 0


Therefore x = 3 and x = 11



Question 18.

Solve the following equations using the method of factorization:

6x2 + 13x + 6 = 0


Answer:

6x2 + 13x + 6 = 0


⇒ 6x2 + 9x + 4x + 6 = 0


taking 3x common from first two terms and 2 common from next two


⇒ 3x(2x + 3) + 2(2x + 3) = 0


⇒ (3x + 2)(2x + 3) = 0


⇒ 3x + 2 = 0 or 2x + 3 = 0


⇒ 3x = – 2 or 2x = – 3


Therefore  and 



Question 19.

Solve the following equations using the method of factorization:

15x2 — 16x + 1 = 0


Answer:

15x2 — 16x + 1 = 0


⇒ 15x2 – 15x– x + 1 = 0


taking 15x common from first two terms and – 1 common from next two


⇒ 15x(x – 1) – 1(x – 1) = 0


⇒ (15x – 1)(x – 1) = 0


⇒ 15x – 1 = 0 or x – 1 = 0


⇒ 15x = 1 or x = 1


Therefore  and x = 1



Question 20.

Solve the following equations using the method of factorization:



Answer:

√5x2 – 4x – √5 = 0


⇒ √5x2 – 5x + x – √5 = 0


taking √5x common from first two terms and 1 common from next two


⇒ √5x(x – √5) + 1(x – √5) = 0


⇒ (√5x + 1)(x – √5) = 0


⇒ √5x + 1 = 0 or x – √5 = 0


⇒ √5x = – 1 or x = √5


Therefore  and x = √5



Question 21.

Solve the following equations using the method of factorization:



Answer:





⇒ 6(x2 + 1) = 13x


⇒ 6x2 + 6 = 13x


⇒ 6x2 – 13x + 6 = 0


⇒ 6x2 – 9x– 4x + 6 = 0


taking 3x common from first two terms and – 2 common from next two


⇒ 3x(2x – 3) – 2(2x – 3) = 0


⇒ (3x – 2)(2x – 3) = 0


⇒ 3x – 2 = 0 or 2x – 3 = 0


⇒ 3x = 2 or 2x = 3


Therefore  and 




Exercise 4.2
Question 1.

Find the discriminant of the following quadratic equations and discuss the nature of the roots:

6x2 — 13x + 6 = 0


Answer:

Comparing equation 6x2 — 13x + 6 = 0 with ax2 + bx + c = 0 we get


a = 6, b = – 13 and c = 6


Discriminant (D) = b2 – 4ac


⇒ D = (– 13)2 – 4(6)(6)


⇒ D = 169 – 4 × 36


⇒ D = 169 – 144


⇒ D = 25


As D > 0 roots of equation 6x2 — 13x + 6 = 0 are real and distinct



Question 2.

Find the discriminant of the following quadratic equations and discuss the nature of the roots:

√6 x2 — 5x + √6 = 0


Answer:

√6x2 – 5x + √6 = 0


Comparing equation √6x2 – 5x + √6 = 0 with ax2 + bx + c = 0 we get


a = √6, b = – 5 and c = √6


Discriminant (D) = b2 – 4ac


⇒ D = (– 5)2 – 4(√6)(√6)


⇒ D = 25 – 4 × 6


⇒ D = 25 – 24


⇒ D = 1


As D > 0 roots of equation √6x2 – 5x + √6 = 0 are real and distinct



Question 3.

Find the discriminant of the following quadratic equations and discuss the nature of the roots:

24x2 — 17x + 3 = 0


Answer:

Comparing equation 24x2 — 17x + 3 = 0 with ax2 + bx + c = 0 we get


a = 24, b = – 17 and c = 3


Discriminant (D) = b2 – 4ac


⇒ D = (– 17)2 – 4(24)(3)


⇒ D = 289 – 4 × 72


⇒ D = 289 – 288


⇒ D = 1


As D > 0 roots of equation 24x2 — 17x + 3 = 0 are real and distinct



Question 4.

Find the discriminant of the following quadratic equations and discuss the nature of the roots:

x2 + 2x + 4 = 0


Answer:

Comparing equation x2 + 2x + 4 = 0 with ax2 + bx + c = 0 we get


a = 1, b = 2 and c = 4


Discriminant (D) = b2 – 4ac


⇒ D = 22 – 4(1)(4)


⇒ D = 4 – 16


⇒ D = – 12


As D < 0 equation x2 + 2x + 4 = 0 has no real solution



Question 5.

Find the discriminant of the following quadratic equations and discuss the nature of the roots:

x2 + x + 1 = 0


Answer:

Comparing equation x2 + x + 1 = 0 with ax2 + bx + c = 0 we get


a = 1, b = 1 and c = 1


Discriminant (D) = b2 – 4ac


⇒ D = 12 – 4(1)(1)


⇒ D = 1 – 4


⇒ D = – 3


As D < 0 equation x2 + x + 1 = 0 has no real solution



Question 6.

Find the discriminant of the following quadratic equations and discuss the nature of the roots:

x2 — 3√3x — 30 = 0


Answer:

Comparing equation x2 — 3√3x — 30 = 0 with ax2 + bx + c = 0 we get


a = 1, b = – 3√3 and c = – 30


Discriminant (D) = b2 – 4ac


⇒ D = (– 3√3)2 – 4(1)(– 30)


⇒ D = (32)(√3)2 + 120


⇒ D = 9 × 3 + 120


⇒ D = 27 + 120


⇒ D = 147


As D > 0 roots of equation x2 — 3√3x — 30 = 0 are real and distinct



Question 7.

If a, b, c  R, a > 0, c < 0, then prove that the roots of ax2 + bx + c = 0 are real and distinct.


Answer:

For roots to be real and distinct D should be greater than 0


Discriminant (D) = b2 – 4ac


In the discriminant b2 is a positive number since square cannot be negative given a > 0 which means a is also positive


Now consider the product – 4 × c


As c < 0 c is negative


We are multiplying two negative numbers which would result in a positive number which means – 4 × c is also positive


So, we can conclude that b2 – 4ac is positive


Hence D > 0


Hence roots are real and distinct



Question 8.

Find k, if the roots of x2 — (3k — 2)x + 2k = 0 are equal and real.


Answer:

Comparing equation x2 — (3k — 2)x + 2k = 0 with ax2 + bx + c = 0 we get


a = 1, b = – (3k – 2) and c = 2k


Discriminant (D) = b2 – 4ac


As roots are real and equal D = 0


⇒ b2 – 4ac = 0


⇒ [ – (3k – 2)]2 – 4(1)(2k) = 0


⇒ (3k – 2)2 – 8k = 0


Expand using (a – b)2 = a2 – 2ab + b2


⇒ 9k2 – 12k + 4 – 8k = 0


⇒ 9k2 – 18k– 2k + 4 = 0


taking 9k common from first two terms and – 2 common from next two


⇒ 9k(k – 2) – 2(k – 2) = 0


⇒ (9k – 2)(k – 2) = 0


⇒ 9k – 2 = 0 or k – 2 = 0


⇒ 9k = 2 or k = 2


Therefore k =  and k = 2



Question 9.

If the roots of the quadratic equation (k + 1)x2 — 2(k—1)x + 1 = 0 are real and equal, find the value of k.


Answer:

Comparing equation (k + 1)x2 — 2(k—1)x + 1 = 0 with ax2 + bx + c = 0 we get


a = (k + 1), b = – 2(k – 1) and c = 1


Discriminant (D) = b2 – 4ac


As roots are real and equal D = 0


⇒ b2 – 4ac = 0


⇒ [ – 2(k – 1)]2 – 4(k + 1)(1) = 0


⇒ (4)(k – 1)2 – 4k – 4 = 0


Expand (k – 1)2 using (a – b)2 = a2 – 2ab + b2


⇒ 4(k2 – 2k + 1) – 4k – 4 = 0


⇒ 4k2 – 8k – 4k = 0


⇒ 4k2 – 12k = 0


Take 4k factor common


⇒ 4k(k – 3) = 0


⇒ 4k = 0 or k – 3 = 0


⇒ k = 0 or k = 3


Therefore k = 0 and k = 3



Question 10.

If the roots of ax2 + 2bx + c = 0, a ≠ 0, a, b, c ∈ R are real and equal, then prove that a : b = b : c.


Answer:

(To avoid confusion between variables here I have taken standard from as px2 + qx + r = 0 and accordingly the discriminant)


Comparing equation ax2 + 2bx + c = 0 with standard form px2 + qx + r = 0 we get


p = a, q = 2b and r = c


Discriminant (D) = q2 – 4pq


As roots are real and equal D = 0


⇒ q2 – 4pq = 0


⇒ (2b)2 – 4(a)(c) = 0


⇒ 4b2 – 4ac = 0


⇒ 4b2 = 4ac


⇒ b2 = ac


⇒ b × b = a × c


⇒  = 


⇒ b:c = a:b


⇒ a:b = b:c


Hence proved



Question 11.

Solve the following equations using the general formula:

x2 + 10x + 6 = 0


Answer:

Comparing equation x2 + 10x + 6 = 0 with ax2 + bx + c = 0 we get


a = 1, b = 10 and c = 6


Discriminant (D) = b2 – 4ac


⇒ D = 102 – 4(1)(6)


⇒ D = 100 – 24


⇒ D = 76


D > 0 implies roots are real and distinct and given by






Therefore x = – 5 + √19 and x = – 5 – √19



Question 12.

Solve the following equations using the general formula:

x2 + 5x — 1 = 0


Answer:

Comparing equation x2 + 5x — 1 = 0 with ax2 + bx + c = 0 we get


a = 1, b = 5 and c = – 1


Discriminant (D) = b2 – 4ac


⇒ D = 52 – 4(1)(– 1)


⇒ D = 25 + 4


⇒ D = 29


D > 0 implies roots are real and distinct and given by




Therefore  and 



Question 13.

Solve the following equations using the general formula:

x2 — 3x — 2 = 0


Answer:

Comparing equation x2 — 3x — 2 = 0 with ax2 + bx + c = 0 we get


a = 1, b = – 3 and c = – 2


Discriminant (D) = b2 – 4ac


⇒ D = (– 3)2 – 4(1)(– 2)


⇒ D = 9 + 8


⇒ D = 17


D > 0 implies roots are real and distinct and given by




Therefore  and 



Question 14.

Solve the following equations using the general formula:

x2 — 3√6x + 12 = 0


Answer:

Comparing equation x2 — 3√6x + 12 = 0 with ax2 + bx + c = 0 we get


a = 1, b = – 3√6 and c = 12


Discriminant (D) = b2 – 4ac


⇒ D = (– 3√6)2 – 4(1)(12)


⇒ D = (32)(√6)2 – 48


⇒ D = 9 × 6 – 48


⇒ D = 54 – 48


⇒ D = 6


D > 0 implies roots are real and distinct and given by




and 


 and 


Therefore x = 2√6 and x = √6



Question 15.

Solve the following equations using the general formula:

3x2 + 5√2x + 2 = 0


Answer:

Comparing equation 3x2 + 5√2x + 2 = 0 with ax2 + bx + c = 0 we get


a = 3, b = 5√2 and c = 2


Discriminant (D) = b2 – 4ac


⇒ D = (5√2)2 – 4(3)(2)


⇒ D = (52)(√2)2 – 4 × 6


⇒ D = 25 × 2 – 24


⇒ D = 50 – 24


⇒ D = 26


D > 0 implies roots are real and distinct and given by







Therefore and



Question 16.

Solve the following equations using the general formula:



Answer:

Cross multiply


⇒ 5 × (x2 – 1) = 4 × (x2 + 1)


⇒ 5x2 – 5 = 4x2 + 4


⇒ 5x2 – 5 – 4x2 – 4 = 0


⇒ x2 – 9 = 0


⇒ x2 + 0x – 9 = 0


Comparing equation x2 + 0x – 9 = 0 with ax2 + bx + c = 0 we get


a = 1, b = 0 and c = – 9


Discriminant (D) = b2 – 4ac


⇒ D = 02 – 4(1)(– 9)


⇒ D = 36


D > 0 implies roots are real and distinct and given by




and 


Therefore x = 3 and x = – 3




Exercise 4.3
Question 1.

Find two numbers whose sum is 27 and the product is 182.


Answer:

Let the two numbers be x and 27 – x so that their sum is 27


Product = 182


⇒ x(27 – x) = 182


⇒ 27x – x2 = 182


⇒ x2 – 27x + 182 = 0


⇒ x2 – 14x– 13x + 182 = 0


taking x common from first two terms and – 13 common from next two


⇒ x(x – 14) – 13(x – 14) = 0


⇒ (x – 13)(x – 14) = 0


⇒ (x – 13) = 0 or (x – 14) = 0


Therefore x = 13 and x = 14


Take any value of x 13 or 14 if we subtract it from 27 we get the other value 13/14


Hence the numbers are 13 and 14 whose sum is 27 and product is 182



Question 2.

Find two consecutive natural numbers, sum of whose squares is 365.


Answer:

Let the two consecutive numbers be x and (x + 1)


Sum of squares of these numbers is 365


⇒ x2 + (x + 1)2 = 365


⇒ x2 + x2 + 2x + 1 = 365


⇒ 2x2 + 2x + 1 – 365 = 0


⇒ 2x2 + 2x – 364 = 0


Take 2 as common factor


⇒ 2 × (x2 + x – 182) = 0


⇒ x2 + x – 182 = 0


⇒ x2 + 14x – 13x – 182 = 0


taking x common from first two terms and – 13 common from next two


⇒ x(x + 14) – 13(x + 14) = 0


⇒ (x – 13)(x + 14) = 0


⇒ (x – 13) = 0 or (x + 14) = 0


Therefore x = 13 and x = – 14


if we take x = 13 then the two consecutive numbers whose sum of squares is 365 will be 13 and 14


if we take x = – 14 then the two consecutive numbers whose sum of squares is 365 will be – 13 and – 14



Question 3.

The sum of ages of two friends is 20 years. Four years ago the product of their ages was 48. Show that these statements cannot be true.


Answer:

Let age of one friend be x and another friend be (20 – x) so that their sum is 20


Age four years ago


Age of one friend would have been (x – 4) and another friend would have been (20 – x – 4) which is (16 – x)


Given that four years ago the product of their ages was 48


⇒ (x – 4)(16 – x) = 48


⇒ 16x – x2 – 64 + 4x = 48


⇒ – x2 – 64 + 20x = 48


⇒ x2 + 64 – 20x + 48 = 0


⇒ x2 – 20x + 112 = 0


Comparing equation x2 – 20x + 112 = 0 with ax2 + bx + c = 0 we get


a = 1, b = – 20 and c = 112


Discriminant (D) = b2 – 4ac


⇒ D = (– 20)2 – 4(1)(112)


⇒ D = 400 – 448


⇒ D = –48


D<0 which means no real values of x


As no real values of x the equation which we formed using given statements cannot be true


Hence the statements cannot be true



Question 4.

A rectangular garden is designed such that the length of the garden is twice its breadth and the area of the garden is 800 m2. Find the length of the garden.


Answer:

Let the breadth of garden be b


Thus length = 2b


Area of garden = 800 m2


Area = length × breadth


⇒ 800 = 2b × b


⇒ 800 = 2b2



⇒ b2 = 400


⇒ b2 = 202


⇒ b2 – 202 = 0


Using identity (a + b)(a – b) = a2 – b2


⇒ (b + 20)(b – 20) = 0


⇒ (b + 20) = 0 or (b – 20) = 0


Therefore b = 20 and not – 20 because b cannot be negative as b is breadth of rectangle and length or breadth cannot be negative


Breadth = 20


Therefore length = 2 × 20 = 40


Therefore, length of the garden is 40 m



Question 5.

Perimeter of a rectangular garden is 360 m and its area is 8000 m2. Find the length of the garden and also find its breadth. (The length is greater than the breadth)


Answer:

Let l be length of rectangle and b be breadth of rectangular garden


Perimeter = 360 m


Perimeter = 2 × (length + breadth) = 2(l + b)


⇒ 2(l + b) = 360


⇒ l + b = 180


⇒ l = 180 – b …(i)


Area = 8000


Area = length × breadth


⇒ 8000 = lb


Using (i)


⇒ 8000 = (180 – b) × b


⇒ 8000 = 180b – b2


⇒ b2 – 180b + 8000 = 0


⇒ b2 – 100b– 80b + 8000 = 0


taking b common from first two terms and – 80 common from next two


⇒ b2 – 100b – 80b + 8000 = 0


⇒ b(b – 100) – 80(b – 100) = 0


⇒ (b – 80)(b – 100) = 0


⇒ (b – 80) = 0 or (b – 100) = 0


b = 80 m because it is given that breadth is smaller than length


using (i) l = 180 – 80 = 100 m


Therefore, length of garden is 100 m and breadth is 80 m



Question 6.

If a cyclist travels at a speed 2 km/hr more than his usual speed, he reaches the destination 2 hours earlier. If the destination is 35 km away, what is the usual speed of the cyclist?


Answer:

Distance = 35 km


Let the usual speed be x


And t be the time taken to reach destination when speed is x




…(i)


If he travels at a speed (x + 2) km/hour time taken is (t – 2) hours


Distance is same 35 km


Speed = 



Using (i)





⇒ (x + 2)(35 – 2x) = 35x


⇒– 2x2 + 70 – 4x = 0


divide by – 2


⇒ x2 + 2x – 35 = 0


⇒ x2 + 7x– 5x – 35 = 0


taking x common from first two terms and – 5 common from next two


⇒ x(x + 7) – 5(x + 7) = 0


⇒ (x – 5)(x + 7) = 0


⇒ (x – 5) = 0 or (x + 7) = 0


Thus x = 5 and not 7 because x represent speed and speed cannot be negative


The usual speed of cyclist = 5 km/hour



Question 7.

The diagonal of a rectangular ground is 60 meters more than the breadth of the ground. If the length of the ground is 30 meters more than the breadth, find the area of the ground.


Answer:

Let the breadth of the ground be b


Diagonal = b + 60


Length = b + 30



Using Pythagoras theorem


(length)2 + (breadth)2 = (diagonal)2


⇒ (b + 30)2 + b2 = (b + 60)2


⇒ (b + 30)2 – (b + 60)2 = – b2


Using identity (a + b)(a – b) = a2 – b2


⇒ (b + 30 + b + 60)(b + 30 – b – 60) = – b2


⇒ (2b + 90)(– 30) = – b2


⇒ b2 – 60b – 2700 = 0


⇒ b2 – 90b + 30b – 2700 = 0


taking b common from first two terms and 30 common from next two


⇒ b(b – 90) + 30(b – 90) = 0


⇒ (b + 30)(b – 90) = 0


⇒ (b + 30) = 0 or (b – 90) = 0


Thus b = 90 m as b cannot be negative because b represents breadth of rectangle


Length = b + 30 = 90 + 30 = 120 m


Area = length × breadth = 120 × 90 = 10800 m2


Therefore, area of ground is 10800 m2



Question 8.

The sides of a right angled triangle are consecutive positive integers. Find the area of the triangle.


Answer:

Let the positive integers x, (x + 1) and (x + 2) be sides of right angled triangle


As (x + 2) will be the greatest number so (x + 2) is the hypotenuse



Using Pythagoras theorem


⇒ x2 + (x + 1)2 = (x + 2)2


⇒ x2 = (x + 2)2 – (x + 1)2


Using identity (a + b)(a – b) = a2 – b2


⇒ x2 = (x + 2 + x + 1)(x + 2 – x – 1)


⇒ x2 = (2x + 3)(1)


⇒ x2 = 2x + 3


⇒ x2 – 2x – 3 = 0


⇒ x2 – 3x + x – 3 = 0


taking x common from first two terms and 1 common from next two


⇒ x(x – 3) + 1(x – 3) = 0


⇒ (x + 1)(x – 3) = 0


⇒ (x + 1) = 0 or (x – 3) = 0


Thus x = 3 because x cannot be negative since x represent he side of a triangle and side cannot be a negative quantity


x + 1 = 3 + 1 = 4


x + 2 = 3 + 2 = 5


Thus, the three sides are 3, 4 and 5


As it is a right angled triangle one side would be base and the other height


Base = 3 and height = 4


Area of triangle =  × base × height


Area of triangle =  × 3 × 4 =  × 12


Therefore, area of triangle is 6 unit2




Exercise 4
Question 1.

Solve the following quadratic equations using factorization:

x2 — 12 = 0


Answer:

(x – 2√ 3)(x + 2√ 3) = 0

(∵ a2 – b2 = (a + b)(a – b), and 12 = (2√ 3)2)


⇒ x – 2√ 3 = 0 or x + 2√ 3 = 0


⇒ x = 2√ 3 or x = – 2√ 3



Question 2.

Solve the following quadratic equations using factorization:

x2 — 7x — 60 = 0


Answer:

x2 – 7x – 60 = 0

⇒ x2 – (12 – 5)x – 60 = 0


⇒ x2 – 12x + 5x – 60 = 0


⇒ x(x – 12) + 5(x – 12) = 0


⇒ (x – 12)(x + 5) = 0


⇒ x – 12 = 0 or x + 5 = 0


⇒ x = 12 or x = – 5



Question 3.

Solve the following quadratic equations using factorization:

x2— 15x + 56 = 0


Answer:

x2— (8 + 7)x + 56 = 0

⇒ x2 – 8x – 7x + 56 = 0


⇒ x(x – 8) – 7(x – 8) = 0


⇒ (x – 7)(x – 8) = 0


⇒ x – 7 = 0 or x – 8 = 0


⇒ x = 7 or x = 8



Question 4.

Solve the following quadratic equations using factorization:



Answer:






(∵ (a2 + b2) + (a2 – b2) = 2(a2 + b2)



⇒ 4(8x2 + 18) = 17(4x2 – 9)


⇒ 32x2 + 72 = 68x2 – 153


⇒ 68x2 – 32x2 = 153 – 72


⇒ 36x2 = 81






Question 5.

Solve the following quadratic equations using factorization:



Answer:





⇒ (15x + 19)(x + 2) = (12x2 + 64x + 20)


⇒ 15x2 + 19x + 30x + 38 = 12x2 + 64x + 20


⇒ 15x2 + 49x + 38 = 12x2 + 64x + 20


⇒ 15x2 – 12x2 + 49x – 64x + 38 – 20 = 0


⇒ 3x2 – 15x + 18 = 0


⇒ 3x2 – (18 – 3)x + 18 = 0


⇒ 3x2 – 18x – 3x + 18 = 0


⇒ 3x(x – 6) – 3(x – 6) = 0


⇒ (3x – 3)(x – 6) = 0


⇒ x = 1 or x = 6.



Question 6.

Find the roots of the following equations by the method of perfect square:

x2 — 24x — 16 = 0


Answer:

x2 — 24x + 144 – 160 = 0


⇒ (x – 12)2 – 160 = 0


⇒ (x – 12)2 – (4√ 10)2 = 0


⇒ (x – 12 – 4√ 10)(x – 12 + 4√ 10) = 0


⇒ x – 12 – 4√ 10 = 0 or x – 12 + 4√ 10 = 0


⇒ x = 12 + 4√ 10 or x = 12 – 4√ 10



Question 7.

Find the roots of the following equations by the method of perfect square:

3x2 + 7x — 20 = 0


Answer:

Multiplying the whole equation by 12, we get –

⇒ 36x2 + 84x — 240 = 0


⇒ 36x2 + 84x + 49 – 289 = 0



⇒ (6x + 7)2 – 172 = 0


⇒ (6x + 7 – 17)(6x + 7 + 17) = 0


⇒ (6x – 10)(6x + 24) = 0


⇒6x – 10 = 0 or 6x + 24 = 0





Question 8.

Find the roots of the following equations by the method of perfect square:

x2 – 10x + 25 = 0


Answer:

x2— 2(1)(5)x + 52 = 0

⇒ (x – 5)2 = 0


⇒ (x – 5)(x – 5) = 0


⇒ x – 5 = 0 or x – 5 = 0


⇒ x = 5



Question 9.

Find the roots of the following equations by the method of perfect square:

x2 + (x + 5)2 = 625


Answer:

x2 + x2 + 10x + 25 – 625 = 0

⇒ 2x2 + 10x – 600 = 0


Dividing the whole equation by 2, we get,


⇒ x2 + 5x – 300 = 0






⇒(x + 20) = 0 or (x – 15) = 0


⇒ x = – 20 or x = 15



Question 10.

Find the roots of the following equations by the method of perfect square:

(x + 2)(x + 3) = 240


Answer:

(x + 2)(x + 3) = 240

⇒ x(x + 3) + 2(x + 3) = 240

⇒ x2 + 2x + 3x + 6 = 240


⇒ x2 + 5x – 234 = 0

Now we have to factorize this quadratic equation by perfect square method

Now for forming a perfect square certain steps should be taken

(i) make the first term's coefficient as 1 and apply the formula for third term)




(ii) Take both the squares so formed and apply the formula a2 - b2 = (a + b) (a - b)


Simplifying we get,

⇒(x – 18) = 0 or (x + 13) = 0


⇒ x = 18 or x = – 13 Answer..


Question 11.

Divide 20 into two parts such that the sum of the square of the parts is 218.


Answer:

Let the divided two parts be x and 20 – x.

⇒ x2 + (20 – x)2 = 218


⇒ x2 + 400 – 40x + x2 = 218


⇒ 2x2 – 40x + 182 = 0


Dividing the whole equation by 2, we get –


⇒ x2 – 20x + 91 = 0


⇒ x2 – (7 + 13)x + 91 = 0


⇒ x2 – 7x – 13x + 91 = 0


⇒ x(x – 7) – 13(x – 7) = 0


⇒ (x – 7)(x – 13) = 0


⇒ x – 7 = 0 or x – 13 = 0


⇒ x = 7 or x = 13


∴ the two parts are 7 and 13.



Question 12.

A car takes 1 hour less to cover a distance of 200 km if its speed is increased by 10 km/hr, than its usual speed. What is the usual speed of the car?


Answer:

Let the usual speed be – x km/hr and time taken to cover 200 km be – t.

⇒ 



Now, when the car speed will be increased by the 10 km/hr, then the speed will become – (x + 10) km/hr.


Now, let the time taken now be T



But given that car took 1 hr. less to cover 200 km by new speed.


⇒ t – T = 1




⇒ 200x + 2000 – 200x = x2 + 10x


⇒ x2 + 10x – 2000 = 0


⇒ x2 + (50 – 40)x – 2000 = 0


⇒ x2 + 50x – 40x – 2000 = 0


⇒ x(x + 50) – 40(x + 50) = 0


⇒ (x + 50)(x – 40) = 0


⇒ x = – 50 or x = 40


And speed cannot be negative.


⇒ x = 40 km/hr.



Question 13.

When there is a decrease of 5 km/hr in the usual uniform speed of a goods train, due to track repair work going on it takes 4 hours more than the usual time for travelling the distance of 400 km. Find the usual speed of the train.


Answer:

Let the usual speed be – x km/hr and time taken to cover 400 km be – t.



Now, when the car speed will be decreased by the 5 km/hr, then the speed will become – (x – 5) km/hr.


Now, let the time taken now be T



But given that car took 4 hr. more to cover 200 km by new speed.


⇒ T = t + 4




⇒ 400x + 2000 – 400x = 4x2 – 20x


⇒ 4x2 – 20x – 2000 = 0


Dividing whole equation by 4, we get –


⇒ x2 – 5x – 500 = 0


⇒ x2 – (25 – 20)x – 2000 = 0


⇒ x2 – 25x + 20x – 2000 = 0


⇒ x(x – 25) + 20(x – 25) = 0


⇒ (x – 25)(x + 20) = 0


⇒ x = 25 or x = – 20


And speed cannot be negative.


⇒ x = 25 km/hr.



Question 14.

A river flows at a speed of 1 km/hr. A boat takes 15 hours to travel 112 km downstream and coming back the same distance upstream. Find the speed of the boat in still water. (Speed of the river flow is less than the speed of the boat in still water)


Answer:

Let speed of boat in still water be x km/hr.

∵ speed of river is given as 1 km/hr


⇒ speed of boat downstream = (x + 1) km/hr.


∴ let time taken to go 112 km downstream = t



Now, speed of boat upstream will be (x – 1) km/hr.


∴ let time taken to go 112 km upstream = T



Given that total journey time was 15 hr


⇒ t + T = 15




⇒ 112x – 112 + 112x + 112 = 15x2 – 15


⇒ 15x2 – 224x – 15 = 0


⇒ 15x2 – 224x – 15 = 0


⇒ 15x2 – (225 – 1)x – 15 = 0


⇒ 15x2 – 225x + x – 15 = 0


⇒ 15x(x – 15) + 1(x – 15) = 0



o r


And speed cannot be negative.


⇒ x = 15 km/hr.



Question 15.

Find a number greater than 1 such that the sum of the number and its reciprocal is 2 .


Answer:

Let the no.be x.





⇒ 15x2 + 15 = 34x


⇒ 15x2 – 34x + 15 = 0


⇒ 15x2 – (9 + 25)x + 15 = 0


⇒ 15x2 – 9x – 25x + 15 = 0


⇒ 3x(5x – 3) – 5(5x – 3) = 0


⇒ (3x – 5) = 0 or (5x – 3 = 0)]]


 or 


 and 


∴ The required no is 



Question 16.

The difference of the speed of a faster car and a slower car is 20 km/hr. If the slower car takes 1 hour more than the faster car to travel a distance of 400 km, find speed of both the cars.


Answer:

Let the usual speed be – x km/hr and time taken to cover 400 km be – t.

⇒ 


Now, when the car speed will be increased by the 20km/hr, then the speed will become – (x + 20) km/hr.


Now, let the time taken now be T



But given that car took 1 hr. less to cover 200 km by new speed.


⇒ t – T = 1




⇒ 400x + 8000 – 400x = x2 + 20x


⇒ x2 + 20x – 8000 = 0


⇒ x2 + (100 – 80)x – 2000 = 0


⇒ x2 + 100x – 80x – 8000 = 0


⇒ x(x + 100) – 80(x + 100) = 0


⇒ (x + 100)(x – 80) = 0


⇒ x = – 1000 or x = 80


And speed cannot be negative.


⇒ x = 80 km/hr.


And speed of faster car = 100 km/hr.



Question 17.

Product of the ages of Virat 7 years ago and 7 years later is 480. Find his present age.


Answer:

Let the present age of Virat be x yrs.

⇒ His age 7 yrs ago was = (x – 7) yrs. and 7 years later will be = (x + 7) yrs.


⇒ (x – 7)(x + 7) = 480


⇒ x2 – 49 = 480


⇒ x2 – 49 – 480 = 0


⇒ x2 – 529 = 0


⇒ x2 – 232 = 0


⇒ (x + 23)(x – 23) = 0


⇒ x + 23 = 0 or x – 23 = 0


⇒ x = 23 or x = – 23


But age cannot be negative


⇒ x = 23 years.



Question 18.

If the age of Sachin 8 year ago is multiplied by his age two years later, the result is 1200. Find the age of Sachin at present.


Answer:

Let present age of Sachin be x yrs.

∴ 8 yrs ago, his age was = (x – 8) yrs.


And, 2 yrs later his age will be = (x + 2) yrs.


⇒ (x – 8)(x + 2) = 1200


⇒ x2 – 6x – 16 = 1200


⇒ x2 – 6x – 1216 = 0


⇒ x2 – (38 – 32)x – 1216 = 0


⇒ x2 – 38x + 32x – 1216 = 0


⇒ x(x – 38) + 32(x – 38) = 0


⇒ (x – 38) + (x + 32) = 0


⇒ x – 38 = 0 or x + 32 = 0


⇒ x = 38 or x = – 32


But age cannot be negative.


⇒ x = 38 years.



Question 19.

Sunita's age at present is 2 years less than 6 times the age of her daughter Anita. The product of their ages 5 years later will be 330. What was the age of Sunita when her daughter Anita was born?


Answer:

Let present age of Anita be x yrs.

∵ sunita’s age is 2 yrs less than 6 times age of Anita.


⇒ Present age of Anita = (6x – 2) yrs.


∴ five yrs later age of Anita will be – (x + 5)yrs.


And age of Sunita = (6x – 2) + 5 yrs


= 6x + 3 years.


⇒ (x + 5)(6x + 3) = 330


⇒ 6x2 + 33x + 15 = 330


⇒6x2 + 33x – 315 = 0


Dividing the whole equation by 3, we get –


⇒2x2 + 11x – 105 = 0


⇒2 + (21 – 10)x – 105 = 0


⇒2x2 + 21x – 10x – 105 = 0


⇒x(2x + 21) – 5(2x + 21) = 0


⇒ (2x + 21)(x – 5) = 0


⇒ 2x + 21 = 0 or x – 5 = 0



But age cannot be negative


⇒ x = 5 years.


∴ present age of Anita = 5 yrs.


And present age of Sunita = 6x – 2


= 6(5) – 2


= 28 years.


And, ∵ Anita was born five years ago,


⇒ Age of Sunita at that time = 28 – 5 yrs.


= 23 yrs.



Question 20.

The formula of the sum of first n natural numbers is S = . If the sum of first n natural number is 325, find n.


Answer:


And, S = 325



⇒ n2 + n = 325(2)


⇒ n2 + n – 650 = 0


⇒ n2 + (26 – 25)n – 650 = 0


⇒ n2 + 26n – 25n – 650 = 0


⇒ n(n + 26) – 25(n + 26) = 0


⇒ (n – 25)(n + 26) = 0


⇒ n – 25 = 0 or n + 26 = 0


⇒ n = 25 or n = – 26


Given that n is a natural no.


⇒ n cannot be negative.


⇒ n = 25



Question 21.

Hypotenuse of a right angled triangle is 2 less than 3 times its shortest side. If the remaining side is 2 more than twice the shortest side, find the area of the triangle.


Answer:

Let length of shortest side(base) be x.

⇒ hypotenuse = 3x – 2.


⇒ remaining side(perpendicular) = 2x + 2



Now, According to Pythagoras theorem,


Hypotenuse2 = Base2 + Perpendicular2


⇒(3x – 2)2 = x2 + (2x + 2)2


⇒ 9x2 + 4 – 12x = x2 + 4x2 + 4 + 8x


⇒ 4x2 – 20x = 0


Dividing the whole equation by 4, we get –


⇒ x2 – 5x = 0


⇒ x(x – 5) = 0


⇒ x = 0 or x – 5 = 0


But side cannot be zero


⇒ x(Base) = 5


⇒Perpendicular = 2x + 2


= 2(5) + 2


= 12


⇒ hypotenuse = 3x – 2


= 3(5) – 2


= 13


Now, area of right triangle –




= 30 sq. unit.



Question 22.

The sum of the squares of two consecutive odd positive integers is 290. Find the numbers.


Answer:

Let first odd positive integer be x.

⇒ second odd positive integer = x + 2


⇒ x2 + (x + 2)2 = 290


⇒ x2 + x2 + 4 + 4x = 290


⇒ 2x2 + 4x – 286 = 0


Dividing the whole equation by 2, we get –


⇒ x2 + 2x – 143 = 0


⇒ x2 + (13 – 11)x – 143 = 0


⇒ x2 + 13x – 11x – 143 = 0


⇒ x(x + 13) – 11(x + 13) = 0


⇒ (x – 11)(x + 13) = 0


⇒ x = 11 or x = – 13


But x is a positive integer


⇒ x = 11 is the first odd positive integer.


And, second consecutive odd positive integer = x + 2


= 13



Question 23.

The product of two consecutive even natural numbers is 224. Find the numbers.


Answer:

Let two consecutive even natural numbers be x and x + 2.

⇒ x(x + 2) = 224


⇒ x2 + 2x – 224 = 0


⇒ x2 + (16 – 14)x – 224 = 0


⇒ x2 + 16x – 14x – 224 = 0


⇒ x(x + 16) – 14(x + 16) = 0


⇒ (x – 14)(x + 16) = 0


⇒ x = 14 or x = – 16


But x is a natural no.


⇒ x = 14 is the first natural no.


And, second no. = x + 2


= 16



Question 24.

The product of digits of a two – digit number is 8 and the sum of the squares of the digits is 20. If the number is less than 25. Find the number.


Answer:

The product of digits of a two – digit number is 8

Let digit at tens place be x




Now,given –




⇒ x4 + 64 = 20x2


⇒ x4 – 20x2 + 64 = 0


⇒ x4 – (16 + 4)x2 + 64 = 0


⇒ x2(x2 – 4) – 16(x2 – 4) = 0


⇒ (x2 – 16) (x2 – 4) = 0


⇒ x = 4 or x = 2


For x = 4,


Original no.


= 42


For x = 2,


Original no. 


= 24


Given no. is less than 25 –


⇒ two – digit no. = 24



Question 25.

If price of sugar decreases by 5, one can buy 1 kg more sugar in ! 150, what is the price of the sugar?


Answer:

Let price of 1 kg sugar be x.

Quantity of sugar 


If price decreased by Rs. 5,then new price = x – 5


∴ quantity can be 


Given, one can buy 1 kg more sugar –



⇒ 150x – 150x + 750 = x2 – 5x


⇒ x2 – 5x – 750 = 0


⇒ x2 – (30 – 25)x – 750 = 0


⇒ x2 – 30x + 25x – 750 = 0


⇒ x(x – 30) + 25(x – 30) = 0


⇒ (x – 30)(x + 25) = 0


⇒ x = 30 or x = – 25


But price cannot be negative.


⇒ x = 30/kg



Question 26.

If the price of petrol is increased by Rs. 5 per litre. One gets 2 litres less petrol spending Rs. 1320. What is the increased price of the petrol?


Answer:

Let original price of 1 ltr. Petrol bex.


If price increased by Rs.5 per ltr.


⇒ New price = x + 5 per ltr.


Given, one gets 2 ltr. Petrol for Rs.1320.



⇒ 1320(x + 5) – 1320x = 2x(x + 5)


⇒ 1320x + 6600x – 1320x – 2 x2 + 10x = 0


⇒ 2x2 + 10x – 6600 = 0


Dividing the equation by 2, we get –


⇒x2 + 5x – 3300 = 0


⇒x2 + (60 – 55)x – 3300 = 0


⇒x2 + 60x – 55x – 3300 = 0


⇒ x(x + 60) – 55(x + 60) = 0


⇒ x = – 60 or x = 55


But price cannot be negative.


⇒ x = Rs.55/ltr.


And increased price = 60/ltr.



Question 27.

A vendor gets a profit in percentage equal to the cost price of a flower pot when he sells it for 96. Find the cost of the flower pot and the percentage of profit.


Answer:

let C.P of flowerpot be x.

⇒ Profit = x%


If C.P = 100, then profit = x.


And, if C.P = x, then Profit –



We know, C.P + Profit = S.P



⇒ 100x + x2 = 9600


⇒ x2 + 100x – 9600 = 0


⇒ x2 + (160 – 60)x – 9600 = 0


⇒ x2 + 160x – 60x – 9600 = 0


⇒ x(x + 160) – 60(x + 160) = 0


⇒ (x – 60)(x + 160) = 0


⇒ x – 60 = 0 or x + 160 = 0


⇒ x = 60 or x = – 160


But price cannot be negative.


⇒ x(C.P) = 60


And profit obtained = 60%



Question 28.

While selling a pen for Rs. 24 the loss in percentage is equal to its cost price. Find the cost price of pen. The cost price of pen is less than 50.


Answer:

Let C.P of pen be x

⇒ Loss = x%


If C.P = Rs.100, then loss = Rs. X


And, if C.P = x, then



Also, Loss = C.P – S.P



⇒ x2 – 100x + 2400 = 0


⇒ x2 – (60 + 40)x + 2400 = 0


⇒ x2 – 60x – 40x + 2400 = 0


⇒ x – 60 = 0 or x – 40 = 0


⇒ x = 60 or x = 40


But given C.P is less than 50


⇒ Cost Price(C.P) = Rs.40



Question 29.

The difference of lengths of sides forming right angle in right angled triangle is 3 cm. If the perimeter of the triangle is 36 cm. Find the area of the triangle.


Answer:

Let the length of perpendicular be x.

Now, given that the difference of perpendicular and base = 3


⇒ Base = x + 3


Also, Perimeter = 36


⇒ 36 = x + (x + 3) + Hypotenuse


⇒ Hypotenuse = 36 – x – x – 3


⇒ 33 – 2x



Now, using Pythagoras Theorem,


Hypotenuse2 = Base2 + Perpendicular2


⇒ (33 – 2x)2 = x2 + (x + 3)2


⇒ 1089 + 4x2 – 132x = x2 + x2 + 9 + 6x


⇒ 2x2 – 1138x + 1080 = 0


Dividing whole equation by 2, we get –


⇒ x2 – 69x + 540 = 0


⇒ x2 – (60 + 9)x + 540 = 0


⇒ x2 – 60x – 9x + 540 = 0


⇒ x(x – 60) – 9(x – 60) = 0


⇒ (x – 60)(x – 9) = 0


⇒ x = 60 or x = 9


∵ Perimeter is given to be 36 cm


⇒ x(Perpendicular) = 9 cm


⇒ Base = x + 3


= 12 cm


Now,




⇒ 54 cm2



Question 30.

The sides of a right angled triangle are x, x + 3, x + 6, x being a positive integer. Find the perimeter of the triangle.


Answer: 


According to Pythagoras theorem,


Hypotenuse2 = Base2 + Perpendicular2


⇒ (x + 6)2 = x2 + (x + 3)2


⇒ x2 + 36 + 12x = x2 + x2 + 9 + 6x


⇒ x2 – 6x – 27 = 0


⇒ x2 – (9 – 3)x – 27 = 0


⇒ x2 – 9x + 3x – 27 = 0


⇒ x(x – 9) + 3(x – 9) = 0


⇒ (x – 9)(x + 3) = 0


⇒ x = 9 or x = – 3


But side can’t be negative.


⇒ x(Base) = 9 cm.


⇒ Perpendicular = x + 3


= 12 cm


And, hypotenuse = x + 6


= 15 cm


∴ the Perimeter = sum of all 3 sides


= 9 + 12 + 15


= 36 cm



Question 31.

…………… is a solution of quadratic equation x2 — 3x + 2 = 0
A. —3

B. 1

C. 3

D. —2


Answer:

x2 — 3x + 2 = 0

⇒ x2 —(2 + 1)x + 2 = 0


⇒ x2 — 2x – x + 2 = 0


⇒ x(x – 2) – 1(x – 2) = 0


⇒ (x – 1) = 0 or (x – 2) = 0


⇒ x = 1 or x = 2


(B) doesn’t match the solution.


(C) doesn’t match the solution.


(D) doesn’t match the solution.


Question 32.

Discriminant D = ….for the quadratic equation 5x2 — 6x + 1 = 0
A. 16

B. 

C. 4

D. 56


Answer:

We know,

D = b2 – 4ac


Here, a = 5, b = – 6 and c = 1


⇒ D = (– 6)2 – 4(5)(1)


⇒ D = 36 – 20


⇒ D = 16


(B) doesn’t match the solution.


(C) doesn’t match the solution.


(D) doesn’t match the solution.


Question 33.

If x = 2 is a root of the equation x2 — 4x + a = 0, then a =
A. —2

B. 2

C. —4

D. 4


Answer:

∵ 2 is a root

⇒ (2)2 – 4(2) + a = 0


⇒ 4 – 8 + a = 0


⇒ a = 4


(A) doesn’t match the solution.


(B) doesn’t match the solution.


(C) doesn’t match the solution.


Question 34.

A quadratic equation has two equal roots, if…….
A. D < 0

B. D > 0

C. D = 0

D. D is non – zero perfect square


Answer:

when D = 0, the quadratic equation has equal roots.

(A) doesn’t match the solution.


(B) doesn’t match the solution.


(C) doesn’t match the solution.


Question 35.

The quadratic equation has 3 as one of its roots.
A. x2 — x — 6 = 0

B. x2 + x — 6 = 0

C. x2 — x + 6 = 0

D. x2 + x + 6 = 0


Answer:

We’ll put x = 3 in all equation and will see if that satisfies.

For (A), x2 — x — 6 = 0


⇒ 9 – 3 – 6 = 0 satisfied.


(B) doesn’t match the solution.


(C) doesn’t match the solution.


(D) doesn’t match the solution.


Question 36.

If 4 is a root of quadratic equation x2 + ax — 8 = 0, then a =
A. 2

B. 4

C. —2

D. —4


Answer:

∵ 4 is a root –

⇒ 42 + a(4) – 8 = 0


⇒ 16 + 4a – 8 = 0


⇒ a = – 2


(A) doesn’t match the solution.


(B) doesn’t match the solution.


(D) doesn’t match the solution.


Question 37.

If one of the roots of kx2 — 7x + 3 = 0 is 3, then k =
A. —2

B. 3

C. —3

D. 2


Answer:

∵ 3 is a root

⇒ k(3)2 – 7(3) + 3 = 0


⇒ 9k – 21 + 3 = 0


⇒ 9k = 18



⇒ k = 2


(A) doesn’t match the solution.


(B) doesn’t match the solution.


(C) doesn’t match the solution.


Question 38.

The discriminate of x2 — 3x — k = 0 is 1. A value of x is
A. —4

B. —2

C. 2

D. 4


Answer:

We know,

D = b2 – 4ac


Here, a = 1, b = – 3 and c = – k


⇒ D = (– 3)2 – 4(1)(– k) = 1


⇒ 9 + 4k = 1


⇒ 4k = – 8



⇒ k = – 2


(A) doesn’t match the solution.


(C) doesn’t match the solution.


(D) doesn’t match the solution.


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