Question

Expand.

(i) ( 5a + 6b)² (ii) ${\left(\frac{a}{2}+\frac{b}{3}\right)}^2$ (iii) ${\left(2p-3q\right)}^2$ (iv) ${\left(x-\frac{2}{x}\right)}^2$

(v) ${\left(ax+by\right)}^2$ (vi) ${\left(7m-4\right)}^2$ (vii) ${\left(x+\frac{1}{2}\right)}^2$ (viii) ${\left(a-\frac{1}{a}\right)}^2$

Answer 1

It is known that, (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².
$$\begin{gathered} \left(\mathrm{i}\right){\left(5a+6b\right)}^2 \\ ={\left(5a\right)}^2+2\times \left(5a\right)\left(6b\right)+{\left(6b\right)}^2 \\ =25a^2+60ab+36b^2 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right){\left(\frac{a}{2}+\frac{b}{3}\right)}^2 \\ ={\left(\frac{a}{2}\right)}^2+2\times \left(\frac{a}{2}\right)\times \left(\frac{b}{3}\right)+{\left(\frac{b}{3}\right)}^2 \\ =\frac{a^2}{4}+\frac{ab}{3}+\frac{b^2}{9} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right){\left(2p-3q\right)}^2 \\ ={\left(2p\right)}^2-2\times \left(2p\right)\times \left(3q\right)+{\left(3q\right)}^2 \\ =4p^2-12pq+9q^2 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right){\left(x-\frac{2}{x}\right)}^2 \\ ={\left(x\right)}^2-2\times \left(x\right)\times \left(\frac{2}{x}\right)+{\left(\frac{2}{x}\right)}^2 \\ =x^2-4+\frac{4}{x^2} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right){\left(ax+by\right)}^2 \\ ={\left(ax\right)}^2+2\times \left(ax\right)\times \left(by\right)+{\left(by\right)}^2 \\ =a^2{x}^2+2axby+b^2{y}^2 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right){\left(7m-4\right)}^2 \\ ={\left(7m\right)}^2-2\times \left(7m\right)\times \left(4\right)+{\left(4\right)}^2 \\ =49m^2-56m+16 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vii}\right){\left(x+\frac{1}{2}\right)}^2 \\ ={\left(x\right)}^2+2\times \left(x\right)\times \left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^2 \\ =x^2+x+\frac{1}{4} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{viii}\right){\left(a-\frac{1}{a}\right)}^2 \\ ={\left(a\right)}^2-2\times \left(a\right)\times \left(\frac{1}{a}\right)+{\left(\frac{1}{a}\right)}^2 \\ =a^2-2+\frac{1}{a^2} \end{gathered}$$