Question

find the value of k for which the equation
$(k - 5) x^{2} + 2(k - 5)x + 2 = 0$
has equal and real roots .
​

Answer 1

Answer:

k = 7

Step-by-step explanation:

Roots are equal

=> discriminant = 0

=> [ 2(k-5) ]² - 4(k-5)(2) = 0

=> (k-5)² - 2(k-5) = 0

=> ( k - 5 ) ( k - 5 - 2 ) = 0

=> ( k - 5 ) ( k - 7 ) = 0

=> k = 5 or k = 7

But for k = 5, the equation becomes 2 = 0, which is absurd.

So we conclude that k = 7 is the only possibility.