Question

If a and b are integers of opposite signs such that $(a+3{)}^2:b^2=9:1$ and $(a-1{)}^2:(b-1{)}^2=4:1$,
then the ratio $a^2:b^2$ is

• A. 9 : 4
• B. 81 : 4
• C. 1 : 4
• D. 25 : 4

Answer 1

The correct option is D

25 : 4

Given, $\frac{(a+3{)}^2}{b^2}=\frac{9}{1}\&\frac{(a-1{)}^2}{(b-1{)}^2}=\frac{4}{1}$
$\frac{(a+3)}{b}=\pm 3\&\frac{(a-1)}{(b-1)}=\pm 2a+3=\pm 3b\dots \dots i)\&a-1=\pm 2(b-1)\Rightarrow a=1\pm 2(b-1)$ putting this value of a in i), we get
$1\pm 2(b-1)+3=\pm 3b4\pm 2(b-1)=\pm 3b\dots \dots ii)$
Clearly the above equation will take 4 different forms based on \pm at two different places. Let's check one by one.

Case 1: Taking '+' sign on both sides of '=' sign
$4+2(b-1)=3b\Rightarrow 4+2b-2=3b\Rightarrow b=2$
Therefore, $a=3b-3=6-3=3$
Since a & b are of opposite signs, this case is not possible as both a & b are positive.

Case 2: Taking '+' sign on left side and '-' sign on right side of '=' sign
4 + 2(b - 1) = - 3b
$2+2b=-3b\Rightarrow 5b=-2\Rightarrow b=-\frac{2}{5}$
Therefore, $a=-3b-3=-3\times \left(\frac{-2}{5}\right)-3=\left(\frac{6}{5}\right)-3=-\frac{9}{5}$
This case is not possible as a & b are of the same sign.

Case 3: Taking '-' sign on left side and '+' sign on right side of '=' sign
$4-2(b-1)=3b\Rightarrow 6-2b=3b\Rightarrow b=\frac{6}{5}$
Therefore, $a=3+3\times \left(\frac{6}{5}\right)=3+\frac{18}{5}=\frac{33}{5}$
This case is not possible as both a & b are of the same sign.

Case 4: Taking '-' sign on left side and '-' sign on right side of '=' sign
$4-2(b-1)=-3b6-2b=-3b\Rightarrow b=-6$
Therefore, $a=-3-3b=-3-3\times (-6)=15$
This is the only possible case as it satisfies the given condition.
Hence, $\frac{a^2}{b^2}=\frac{(15{)}^2}{(-6{)}^2}=\frac{25}{4}$.