Question

The equation $x^2-ax+b=0\&x^3-p{x}^2+qx=0$, where $b\ne 0,q\ne 0$, have one common root & the second equation has two equal roots. Then, $2(q+b)=$.

• A. $2ap$
• B. $a{p}^2$
• C. $a^{2}p$
• D. $ap$

Answer 1

Answer: (D)

$ap$

Let $x^2-ax+b=0$ has root $\alpha$ & $\beta$.
And $x(x^2-px+q)=0$ has root $0,\alpha ,\alpha$.
($x=0$ is not the common root since $b\ne 0$)
Now, $x^2-px+q=0$ has two equal roots.
$\Rightarrow D=0\Rightarrow p^2-4q=0$ & $\frac{p}{2}$ is its root.
$\frac{p}{2}$ satisfies equation $x^2-ax+b=0$
$\frac{p^2}{4}-\frac{ap}{2}+b=0\Rightarrow \frac{p^2}{4}+b=\frac{ap}{2}$
$\Rightarrow 2(q+b)=ap$