Question

Find the value of k for which quadratic equation x^2-(k+2x)+ (-k^2-4k-4)

Answer 1

Step-by-step explanation:

Given that: Find the value of k for which quadraticequation x^2-(k+2)x+ (-k^2-4k-4) has equal roots.

Solution:

We know that a quadratic equation has equal roots, if D= 0

$D= {b}^{2} - 4ac \\ \\ here \\ \\ a = 1 \\ \\ b = - k - 2 \\ \\ c = - {k}^{2} - 4k - 4 \\ \\ D= {( - k - 2)}^{2} - 4(1)( - {k}^{2} - 4k - 4) \\ \\ {(k + 2)}^{2} + 4( {k}^{2} + 4k + 4) = 0 \\ \\ {(k + 2)}^{2} + 4( {k + 2)}^{2} = 0 \\ \\ 5( {k + 2)}^{2} = 0 \\ \\ ( {k + 2)}^{2} = 0 \\ \\ taking \: root \: both \: sides \\ \\ k + 2 = 0 \\ \\ \bold{k = - 2}$

Thus, the value of k is -2,for the quadratic equation have equal zeros.

Hope it helps you.