Question

7. Find the value of K if the equation x2-2(k+1)x + kz = 0 has equal roots.​

Answer 1

Answer:

it is 10 class question of chater quadratic equation

Answer 2

Answer:

Step-by-step explanation:

We know, if there is equal roots then, b^2 = 4az

Now, x^2-(2k+2)x + kz = 0

        => x^2 - (2k + 2 ) + kz = 0

 so, a=1,b= -2k-2, z= k

now, (-2k-2)^2 = 4*1*k

=> 4k^2 + 4 + 8k = 4k

=> 4k^2 + 4 + 8k - 4k = 0

=> 4k^2 + 4 +4k = 0

=> 4 ( k^2 + 1 + k) = 0

=> k^2 + 1 + k = 0

Now, solve this equation to get your answer.