Question

If $\alpha$ and $\beta$ are solutions to the equation $\frac{2y}{2y+1}+\frac{8}{y-3}=0$ and $\alpha <\beta$, what is the value of $\alpha$?

• A. -4
• B. -1
• C. 1
• D. 4

Answer 1

The correct option is A -4

Cross multiplying, we get
$2y(y-3)+8(2y+1)=0$
$\Rightarrow 2y^2-6y+16y+8=0$
$\Rightarrow 2y^2+10y+8=0$
On factoring, we get the roots
$2y^2+2y+8y+8=0$
$\Rightarrow 2y(y+1)+8(y+1)=0$
$\Rightarrow 2y+8=0,y+1=0$
$\Rightarrow 2y=-8$
$\Rightarrow y=-4$
So, $\alpha =-4$