If −4 is a root of the quadratic equation
Question:

If $-4$ is a root of the quadratic equation $x^{2}+2 x+4 p=0$, find the value of $k$ for which the quadratic equation $x^{2}+p x(1+3 k)+7(3+2 k)=0$ has equal

roots.

 

Solution:

It is given that $-4$ is a root of the quadratic equation $x^{2}+2 x+4 p=0$.

$\therefore(-4)^{2}+2 \times(-4)+4 p=0$

$\Rightarrow 16-8+4 p=0$

$\Rightarrow 4 p+8=0$

$\Rightarrow p=-2$

The equation $x^{2}+p x(1+3 k)+7(3+2 k)=0$ has equal roots.

$\therefore D=0$

$\Rightarrow[p(1+3 k)]^{2}-4 \times 1 \times 7(3+2 k)=0$

$\Rightarrow[-2(1+3 k)]^{2}-28(3+2 k)=0$

$\Rightarrow 4\left(1+6 k+9 k^{2}\right)-28(3+2 k)=0$

$\Rightarrow 4\left(1+6 k+9 k^{2}-21-14 k\right)=0$

$\Rightarrow 9 k^{2}-8 k-20=0$

$\Rightarrow 9 k^{2}-18 k+10 k-20=0$

$\Rightarrow 9 k(k-2)+10(k-2)=0$

$\Rightarrow(k-2)(9 k+10)=0$

$\Rightarrow k-2=0$ or $9 k+10=0$

$\Rightarrow k=2$ or $k=-\frac{10}{9}$

Hence, the required value of $k$ is 2 or $-\frac{10}{9}$.

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