Question

i) $(\frac{x}{2}+\frac{3y}{4})(\frac{x}{2}+\frac{3y}{4})$
ii) (7a-9b)(7a-9b)

Answer 1

$i\left)\right(\frac{x}{2}+\frac{3y}{4}\left)\right(\frac{x}{2}+\frac{3y}{4})$

$LHS=(\frac{x}{2}+\frac{3y}{4})(\frac{x}{2}+\frac{3y}{4})=\frac{x^2}{4}+\frac{3xy}{8}+\frac{3xy}{8}+\frac{9y^2}{16}$

$=\frac{4x^2+12xy+9y^2}{16}$

$RHS=(\frac{x}{2}{)}^2+(\frac{3y}{4}{)}^2+2\left(\frac{x}{2}\right)\left(\frac{3y}{4}\right)$

$=\frac{x^2}{4}+\frac{9y^2}{16}+\frac{3xy}{4}$

$=\frac{4x^2+9y^2+12xy}{16}$

LHS = RHS $\therefore$ Hence, proved

$ii\left)\right(7a-9b\left)\right(7a-9b)$

$LHS=(7a-9b)(7a-9b)=49a^2-63ab-63ab+81b^2$

$=49a^2-126ab+81b^2$

RHS $=(7a{)}^2+(9b{)}^2-2\left(7a\right)\left(9b\right)$

$=49a^2+81b^2-12ab$

LHS = RHS

$\therefore$ Hence proved.