The value of $k$ for which the quadratic equation $x^{2} + k (2 x + k + 1) + 4 = 0$ has real and equal roots is

- 2

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4

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2

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- 4

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Solution

The correct option is D $- 4$
Consider the quadratic equation $x^{2} + k (2 x + k + 1) + 4 = 0$
It can be rewritten as $x^{2} + 2 k x + k^{2} + k + 4 = 0$
Now, for real and equal roots, discriminant is zero.
$∴ D = 0$
$\Rightarrow (2 k)^{2} - 4 \times 1 \times (k^{2} + k + 4) = 0$
$\Rightarrow 4 k^{2} - 4 (k^{2} + k + 4) = 0$
$\Rightarrow 4 k^{2} - 4 k^{2} - 4 k - 16 = 0$
$\Rightarrow - 4 k - 16 = 0$
$\Rightarrow k = - 4$
Hence, the correct answer is option (4).

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The value of k for which the quadratic equation x2+k(2x+k+1)+4=0 has real and equal roots is

The value of $k$ for which the quadratic equation $x^{2} + k (2 x + k + 1) + 4 = 0$ has real and equal roots is