 ## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 17 of Exercise 10D

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17
##### 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.

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