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If $(x-2)$ is a common factor of the expressions $\mathrm{x}^2+\mathrm{ax}+\mathrm{b}$ and $\mathrm{x}^2+\mathrm{cx}+\mathrm{d}$, then $\frac{b-d}{c-a}=$
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Verified Answer
The correct answer is:
$2$
$x^2+a x+b$ have $(x-2)$ as factor
$\begin{array}{r}
\therefore \quad(2)^2+2 a+b=0 \\
2 a+b=4 ...(i)
\end{array}$
Similarly, $x^2+c x+d=0$
$\begin{aligned}
& \Rightarrow(2)^2+2 c+d=0 \\
& \Rightarrow 2 c+d=-4... (ii)
\end{aligned}$
(i) - (ii)
$\begin{aligned}
& 2(a-c)+(b-d)=0 \\
\Rightarrow & b-d=2(c-a) \\
\Rightarrow & \frac{b-d}{c-a}=2
\end{aligned}$
$\begin{array}{r}
\therefore \quad(2)^2+2 a+b=0 \\
2 a+b=4 ...(i)
\end{array}$
Similarly, $x^2+c x+d=0$
$\begin{aligned}
& \Rightarrow(2)^2+2 c+d=0 \\
& \Rightarrow 2 c+d=-4... (ii)
\end{aligned}$
(i) - (ii)
$\begin{aligned}
& 2(a-c)+(b-d)=0 \\
\Rightarrow & b-d=2(c-a) \\
\Rightarrow & \frac{b-d}{c-a}=2
\end{aligned}$
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