Question

Find the missing term in the following problem.
${(\frac{3x}{4}-\frac{4y}{3})}^2=\frac{9x^2}{16}+.........+\frac{16y^2}{9}$

• A. $2xy$
• B. $-2xy$
• C. $12xy$
• D. $-12xy$

Answer 1

The correct option is B $-2xy$

The L.H.S. of the algebraic expression is of the form ${(a-b)}^2$

where, $a=\frac{3x}{4}\&b=\frac{4y}{3}$.

${(a-b)}^2=a^2-2ab+b^2⟹(\frac{3x}{4}-\frac{4y}{3})={\left(\frac{3x}{4}\right)}^2-2\times \frac{3x}{4}\times \frac{4y}{3}+{\left(\frac{4y}{3}\right)}^2=\frac{9x^2}{16}-2xy+\frac{16y^2}{9}$.

So the middle term $-2xy$ is missing.