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If $c$ and $d$ are the roots of $x^2+a x+b=0$, then a root of $x^2+(4 c+a) x+\left(b+2 a c+4 c^2\right)=0$ is
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The correct answer is:
$\mathrm{d}-2 \mathrm{c}$
Given that $c$ and $d$ are the roots of $x^2+a x+b=0$
$\begin{aligned}
&\Rightarrow c+d=-a \& c d=b ... (i)\\
&\begin{aligned}
& \text { Now, } x^2+(4 c+a) x+\left(b+2 a c+4 c^2\right)=0 \\
& \Rightarrow x^2(4 c-c-d) x+\left(b-2 c(c+d)+4 c^2\right)=0 \\
& \Rightarrow x^2+(3 c-d) x+2 c^2-c d=0 \\
& \Rightarrow x^2+c x+(2 c-d) x+c(2 c-d)=0 \\
& \Rightarrow(x+c)(x+2 c-d)=0 \\
& \Rightarrow x=d-2 c .
\end{aligned}
\end{aligned}$
$\begin{aligned}
&\Rightarrow c+d=-a \& c d=b ... (i)\\
&\begin{aligned}
& \text { Now, } x^2+(4 c+a) x+\left(b+2 a c+4 c^2\right)=0 \\
& \Rightarrow x^2(4 c-c-d) x+\left(b-2 c(c+d)+4 c^2\right)=0 \\
& \Rightarrow x^2+(3 c-d) x+2 c^2-c d=0 \\
& \Rightarrow x^2+c x+(2 c-d) x+c(2 c-d)=0 \\
& \Rightarrow(x+c)(x+2 c-d)=0 \\
& \Rightarrow x=d-2 c .
\end{aligned}
\end{aligned}$
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