Question

Let $a,b,c,d$ be distinct real numbers such that $a,b$ are roots of $x^2-5cx-6d=0,$ and $c,d$ are roots of $x^2-5ax-6b=0.$ Then $b+d$ is

• A. $180$
• B. $162$
• C. $144$
• D. $126$

Answer 1

The correct option is C $144$

$x^2-5cx-6d=\mathrm{0......}\left(1\right)x^2-5ax-6b=\mathrm{0......}\left(2\right)$

Given that $a,b$ are the roots of the equation $\left(1\right)$ and $c,d$ are the roots of the equation $\left(2\right)$

$\Rightarrow a+b=5c,ab=-6d\&c+d=5a,cd=-6b$

Solving these equation
$b+d=4(a+c).....\left(3\right)d-b=6(a-c).....\left(4\right)$

Now $a$ and $c$ are the roots of equation $\left(1\right)$ and $\left(2\right)$ respectively.
$a^2-5ac-6d=\mathrm{0......}\left(5\right)c^2-5ac-6b=\mathrm{0.......}\left(6\right)$

$\left(5\right)-\left(6\right)\Rightarrow a^2-c^2=6(d-b)\Rightarrow (a+c)(a-c)=6(d-b)\Rightarrow \left(\frac{b+d}{4}\right)\left(\frac{d-b}{6}\right)=6(d-b)(\because \text{eq.}(3\left)\&\right(4\left)\right)\Rightarrow b+d=144$