Question

If $r<0$ and $(4r-4{)}^2=36$, then the value of $r$ is

• A. $-6$
• B. $-2$
• C. $-1$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D $-\frac{1}{2}$

First expand the equation $(4r{)}^2-32r+16-36=0$
$\Rightarrow 16r^2-32r-20=0$
$\Rightarrow 4(4r^2-8r-5)=0$
$\Rightarrow 4r^2-8r-5=0$
On factoring, we get
$4r^2+2r-10r-5=0$
$\Rightarrow 2r(2r+1)-5(2r+1)=0$
$\Rightarrow 2r+1=0,2r-5=0$
$\Rightarrow r=$ $-\frac{1}{2}$ is correct for the condition $r<0$.