Question

Find valies of k for the equation, having equal roots : x^2-2 (k+1)x+k^2.

Answer 1

Finding k value :

Compare x²-2(k+1)x+k²=0

with ax²+bx+c=0 , we get

a=1 , b = -2(k+1) , c = k²

Discreminant (D) = 0

[ Since, roots are equal ]

b²-4ac = 0

=> [2(k+1)]² - 4×1×k² = 0

=> 4(k+1)² = 4k²

=> (k+1)² = k²

=> k+1 = ± k

=> k+1 = - k

=> k+k = -1

=> 2k = -1

=> k = -1/2

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