Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 17 of Exercise 10D
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Find the value of a for which the equation (a – 12)x2 + 2(a – 12)x + 2 = 0 has equal roots.
Given that the quadratic equation (a – 12)x2 + 2(a – 12)x + 2 = 0 has equal roots
Comparing with standard quadratic equation Ax2 + Bx + C = 0
A = (a – 12) B = 2(a – 12) C = 2
Thus D = 0
Discriminant D = B2 – 4AC≥0
[2(a – 12)]2 – 4(a – 12)2≥0
4(a2 + 144 – 24a) – 8a + 96 = 0 using (a – b)2 = a2 – 2ab + b2
4a2 + 576 – 96a – 8a + 96 = 0
4a2 – 104a + 672 = 0
a2 – 26a + 168 = 0
a2 – 14a – 12a + 168 = 0
a(a – 14) – 12(a – 14) = 0
(a – 14)(a – 12) = 0
a = 14 or a = 12
for a = 12 the equation will become non quadratic – – (a – 12)x2 + 2(a – 12)x + 2 = 0
A, B will become zero
Thus value of a = 14 for which the equation has equal roots.
