Question

find the value of k for which the following eq has equal roots (k-12)x^2 +2 (k-12)x+2=0​

Answer 1

Step-by-step explanation:

(k - 12)x² + 2(k - 12)x + 2 = 0

a = k - 12

b = 2(k - 12) = 2k - 24

c = 2

To have equal roots, the discriminant must equal to 0

D = 0

b² - 4ac = 0

(2k - 24)² - 4 ×(k - 12) × 2 = 0

4k² - 96k + 576 - 8k + 96 = 0

4k² - 104k + 672 = 0 (divide by 4 for both sides of equation)

k² - 26k + 168 = 0

k² - 12k - 14k + 168 = 0

k(k - 12) - 14(k - 12) = 0

(k - 12)(k - 14) = 0

k = 12 or k = 14

Note:- But at k= 12, terms of x² and x in the equation vanish hence only k = 14 is possible.

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