Question

If the two equations $x^2-cx+d=0$ and $x^2-ax+b=0$ have a common root and the second equation has equal roots, then

• A. $b+d=ac$
• B. $2(b+d)=ac$
• C. $b+d=2ac$
• D. $(b+d{)}^2=a+c$

Answer 1

The correct option is B $2(b+d)=ac$

Given $x^2-cx+d=0$ $\dots \left(i\right)$

$x^2-ax+b=0$ $\dots \left(ii\right)$

It is given that the second equation has equal roots.
Then
$a^2-4b=0$
Or
$a^2=4b$

Hence equation $\left(ii\right)$ transforms to

$x^2-ax+\frac{a^2}{4}=0$
$x^2-2x\left(\frac{a}{2}\right)+\frac{a^2}{4}=0$
${(x-\frac{a}{2})}^2=0$
$x=\frac{a}{2}$

Hence the common root has to be $\frac{a}{2}$.
Let the common root be $\alpha$

Then equation $\left(i\right),\left(ii\right)$ become

${\alpha }^2-c\alpha +d=0$ ...(ii)
${\alpha }^2-a\alpha +b=0$ ...(iii)

Subtracting (iii) from (ii) gives us

$\alpha (a-c)+d-b=0$
Or
$\alpha =\frac{b-d}{a-c}$

But we know that the common root $\alpha$ is
$=\frac{a}{2}$.
Hence
$\alpha =\frac{b-d}{a-c}$
$\frac{a}{2}=\frac{b-d}{a-c}$
$a(a-c)=2(b-d)$
$a^2-ac=2b-2d$

We know that $a^2=4b$ from i.
Hence
$4b-ac=2b-2d$
$4b-2b+2d=ac$
$2b+2d=ac$
$2(b+d)=ac$