If-4 is a root of the equation $x^{2} + 2 x + 4 p = 0$ , find the value of k for which the quadratic $x^{2} + k (2 x + k + 2) + p = 0$ are equal.

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Solution

The given quadratic equation is
$x^{2} + 2 x + 4 p = 0$
Since −4 is a root of the above equation,
then it must satisfy it.Now,
$(- 4)^{2} + 2 (- 4) + 4 p$
⇒16 − 8 + 4p = 0
⇒4p = −8
⇒p = −2
Now, the other quadratic equation is :

$x^{2} + k (2 x + k + 2) + p = 0$
$x^{2} + k (2 x + k + 2) - 2 = 0$
$x^{2} + 2 k x + k^{2} + 2 k - 2 = 0$
$x^{2} + 2 k x + (k^{2} + 2 k - 2) = 0$

if roots are equal then
D= $b^{2} - 4 a c$ =0
= $(2 k)^{2} - 4 \times 1 \times (k^{2} + 2 k - 2)$
= $4 k^{2} - 4 k^{2} - 8 k + 8$ =0
=8k=8
k=1

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If-4 is a root of the equation x2+2x+4p=0, find the value of k for which the quadratic x2+k(2x+k+2)+p=0 are equal.

If-4 is a root of the equation $x^{2} + 2 x + 4 p = 0$ , find the value of k for which the quadratic $x^{2} + k (2 x + k + 2) + p = 0$ are equal.