### Quadratic Equations Class 10th Mathematics Gujarat Board Solution

##### Question 1.Solve the following quadratic equations using factorization:x2 — 12 = 0Answer:(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√ 3Question 2.Solve the following quadratic equations using factorization:x2 — 7x — 60 = 0Answer: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 = – 5Question 3.Solve the following quadratic equations using factorization:x2— 15x + 56 = 0Answer: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 = 8Question 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 = 0Answer: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√ 10Question 7.Find the roots of the following equations by the method of perfect square:3x2 + 7x — 20 = 0Answer: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 = 0Answer:x2— 2(1)(5)x + 52 = 0⇒ (x – 5)2 = 0⇒ (x – 5)(x – 5) = 0⇒ x – 5 = 0 or x – 5 = 0⇒ x = 5Question 9.Find the roots of the following equations by the method of perfect square:x2 + (x + 5)2 = 625Answer:x2 + x2 + 10x + 25 – 625 = 0⇒ 2x2 + 10x – 600 = 0Dividing the whole equation by 2, we get,⇒ x2 + 5x – 300 = 0    ⇒(x + 20) = 0 or (x – 15) = 0⇒ x = – 20 or x = 15Question 10.Find the roots of the following equations by the method of perfect square:(x + 2)(x + 3) = 240Answer:(x + 2)(x + 3) = 240⇒ x(x + 3) + 2(x + 3) = 240⇒ x2 + 2x + 3x + 6 = 240⇒ x2 + 5x – 234 = 0Now we have to factorize this quadratic equation by perfect square methodNow 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 = 0Dividing 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 = 40And 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 = 0Dividing 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 = – 20And 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 = 80And 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 = – 23But 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 = – 32But 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 = 0Dividing 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 = – 26Given that n is a natural no.⇒ n cannot be negative.⇒ n = 25Question 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 = 0Dividing the whole equation by 4, we get –⇒ x2 – 5x = 0⇒ x(x – 5) = 0⇒ x = 0 or x – 5 = 0But side cannot be zero⇒ x(Base) = 5⇒Perpendicular = 2x + 2= 2(5) + 2= 12⇒ hypotenuse = 3x – 2= 3(5) – 2= 13Now, 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 = 0Dividing 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 = – 13But x is a positive integer⇒ x = 11 is the first odd positive integer.And, second consecutive odd positive integer = x + 2= 13Question 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 = – 16But x is a natural no.⇒ x = 14 is the first natural no.And, second no. = x + 2= 16Question 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 8Let 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 = 2For x = 4,Original no. = 42For x = 2,Original no. = 24Given no. is less than 25 –⇒ two – digit no. = 24Question 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 = – 25But price cannot be negative.⇒ x = 30/kgQuestion 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 = 0Dividing 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 = 55But 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 = – 160But price cannot be negative.⇒ x(C.P) = 60And 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. XAnd, 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 = 40But given C.P is less than 50⇒ Cost Price(C.P) = Rs.40Question 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 + 3Also, 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 = 0Dividing 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 cmNow,  ⇒ 54 cm2Question 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 = – 3But side can’t be negative.⇒ x(Base) = 9 cm.⇒ Perpendicular = x + 3= 12 cmAnd, hypotenuse = x + 6= 15 cm∴ the Perimeter = sum of all 3 sides= 9 + 12 + 15= 36 cmQuestion 31.…………… is a solution of quadratic equation x2 — 3x + 2 = 0A. —3B. 1C. 3D. —2Answer: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 = 0A. 16B. C. 4D. 56Answer:We know,D = b2 – 4acHere, 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. —2B. 2C. —4D. 4Answer:∵ 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 < 0B. D > 0C. D = 0D. D is non – zero perfect squareAnswer: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 = 0B. x2 + x — 6 = 0C. x2 — x + 6 = 0D. x2 + x + 6 = 0Answer: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. 2B. 4C. —2D. —4Answer:∵ 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. —2B. 3C. —3D. 2Answer:∵ 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 isA. —4B. —2C. 2D. 4Answer:We know,D = b2 – 4acHere, 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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