### Quadratic Equations Class 10th Mathematics CBSE Solution

##### Question 1.Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:(i) 2x2 – 3x + 5 = 0(ii) (iii) 2x2 – 6x + 3 = 0Answer:There are three types of roots that are possible for a quadratic equation:For a quadratic equation: a x2 + b x + c = 0, we know that its roots is given by the formula (i) Comparing this equation with ax2 + bx + c = 0, we obtain,a = 2, b = −3, c = 5Discriminant D = b2 − 4ac = (− 3)2 − 4 (2) (5) = 9 − 40 = −31As, b2 − 4ac < 0,Therefore, no real root is possible for the given equation.(ii) Comparing this equation with ax2 + bx + c = 0,we obtain,a = 3b = -4√3c = 4 Discriminant = 48 − 48 = 0As, b2 − 4ac = 0,Therefore, real roots exist for the given equation and they are equal to each other.And the roots will be and . Therefore, the roots are and .(iii) Comparing this equation with ax2+bx + c = 0,we obtain,a = 2, b = −6, c = 3Discriminant = b2 − 4ac = (− 6)2 − 4 (2) (3) = 36 − 24 = 12As, b2 − 4ac > 0,Therefore, distinct real roots exist for this equation as follows.    So, the roots are or .Question 2.Find the values of k for each of the following quadratic equations, so that they have two equal roots.(i) 2x2+ kx + 3 = 0(ii) kx (x – 2) + 6 = 0Answer:We know that if an equation ax2 + bx + c = 0 has two equal roots,its discriminant(b2 − 4ac) will be 0.(i) 2x2 + kx + 3 = 0Comparing equation with ax2 + bx + c = 0, we obtain,a = 2, b = k, c = 3Discriminate = b2 − 4ac = (k)2− 4(2) (3) = k2 − 24For equal roots,Discriminant = 0k2 − 24 = 0k2 = 24= (ii) kx (x − 2) + 6 = 0or kx2− 2kx + 6 = 0Comparing this equation with ax2 + bx + c = 0, we obtain,a = k, b = −2 k, c = 6Discriminant = b2 − 4ac = (− 2k)2 − 4 (k) (6) = 4k2 − 24kFor equal roots, b2 − 4ac = 0= 4k2 − 24k = 0= 4k (k − 6) = 0Either 4k = 0 or k = 6= k = 0 or k = 6However, if k = 0, then the equation will not have the terms ‘x2’ and ‘x’.Therefore, if this equation has two equal roots, k should be 6 only.Question 3.Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.Answer:Let the breadth of mango grove be l.Length of mango grove will be 2l.Area of mango grove = (2l) (l) = 2l22l2 = 800=⇒ ⇒ l2 = 400 However, length cannot be negative.Therefore, breadth of mango grove = 20 mLength of mango grove = 2 × 20 = 40 mQuestion 4.Is the following situation possible? If so, determine their present ages.The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.Answer:Let the age of one friend be x years.Age of the other friend will be (20 − x) years.4 years ago,age of 1st friend = (x − 4) yearsAnd, age of 2nd friend = (20 − x − 4) = (16 − x) yearsGiven that,(x − 4) (16 − x) = 4816x − 64 − x2 + 4x = 48x2 − 20x + 112 = 0Comparing this equation with ax2 + bx + c = 0, we obtaina = 1, b = −20, c = 112Discriminant = b2 − 4ac = (− 20)2 − 4 (1) (112) = 400 − 448 = −48As b2 − 4ac < 0,Therefore, no real root is possible for this equation and hence, this situation is not possible.Question 5.Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.Answer:Let the length and breadth of the park be l and b.Perimeter = 2 (l + b) = 80l + b = 40 Or, b = 40 – lArea = l × b = l (40 − l)= 40l − l2= 400 Givenl2 − 40l + 400 = 0Comparing this equation with al2 + bl + c = 0, we obtaina = 1, b = −40, c = 400Discriminant D = b2 − 4ac = (− 40)2 −4 (1) (400) = 1600 − 1600 = 0As b2 − 4ac = 0,Therefore, this equation has equal real roots and hence, this situation is possible.Root of this equation,  Therefore, length of park, l = 20 mAnd breadth of park, b = 40 − l = 40 − 20 = 20 m

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