Question : If $ab=21$ and $\frac{(a+b)^{2}}{(a-b)^{2}}=\frac{25}{4}$, then the value of $a^{2}+b^{2}+3ab$ is:
Option 1: 115
Option 2: 121
Option 3: 125
Option 4: 127
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Correct Answer: 121
Solution :
$\frac{(a+b)^{2}}{(a-b)^{2}}=\frac{25}{4}$
Taking root on both sides,
⇒ $\frac{(a+b)}{(a-b)}=\frac{5}{2}$
Cross Multiplying both sides:
⇒ $(2a+2b)=(5a-5b)$
⇒ $3a=7b$
⇒ $a=\frac{7}{3}b$
Given: $ab=21$
⇒ $\frac{7}{3}b\times b=21$
⇒ $\frac{7}{3}b^{2}=21$
⇒ $b^{2}=\frac{21\times 3}{7}$
⇒ $b^{2}=9$
Now putting the value in $a^{2}+b^{2}+3ab$
$=\frac{49}{9}b^{2}+9+3\times 21$
$=\frac{49}{9}\times 9+9+3\times 21$
$=49+9+63$
$=121$
Hence, the correct answer is 121.
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