Question

Find the value of expression $81x^2+144y^2-216xy$, when $x=\frac{1}{3}$ and $y=\frac{1}{4}$.

• A. $0$
• B. $6$
• C. $9$
• D. $4$

Answer 1

The correct option is A $0$

Given expression is
$81x^2+144y^2-216xy$

On simplifying we get,
$=(9x{)}^2+(12y{)}^2-2\times \left(9x\right)\times \left(12y\right)$

$=(9x-12y{)}^2$
$[\because a^2+b^2-2ab=(a-b{)}^2]$

Putting values of $x=\frac{1}{3}$ and $y=\frac{1}{4}$ we get,
$=(9\times \frac{1}{3}-12\times \frac{1}{4}{)}^2$

$=(3-3{)}^2$

$=0$