Question

Find the square of :

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

(ii) $\frac{5a}{6b}-\frac{6b}{5a}$

(iii) $2m^2-\frac{2}{3}{n}^2$

(iv) $5x+\frac{1}{5x}$

(v) $8x+\frac{3}{2}y$

(vi) $607$

(vii) $391$

(viii) $9.7$

Answer 1

(i) $3x+\frac{2}{y}{(3x+\frac{2}{y})}^2=(3x{)}^2+{\left(\frac{2}{y}\right)}^2+2(3x)\left(\frac{2}{y}\right)=9x^2+\frac{4}{y^2}+\frac{12x}{y}$

(ii) ${(\frac{5a}{6b}-\frac{6b}{5a})}^2={\left(\frac{5a}{6b}\right)}^2+{\left(\frac{6b}{5a}\right)}^2-2\times \frac{5a}{6b}\times \frac{6b}{5a}=\frac{25a^2}{36b^2}-2+\frac{36b^2}{25a^2}$

(iii) $2m^2-\frac{2}{3}{n}^2{(2m^2-\frac{2}{3}{n}^2)}^2=(2m^2{)}^2+{\left(\frac{2}{3}{n}^2\right)}^2-2\times 2m^2\times \frac{2}{3}{n}^2=4m^4+\frac{4}{9}{n}^4-\frac{8}{3}{m}^2{n}^2=4m^4-\frac{8}{3}{m}^2{n}^2+\frac{4}{9}{n}^2$

(iv) ${(5x+\frac{1}{5x})}^2=(5x{)}^2+\frac{1}{(5x{)}^2}+2\times 5x\times \frac{1}{5x}=25x^2+\frac{1}{25x^2}+2=25x^2+2+\frac{1}{25x^2}$

(v) ${(8x+\frac{3}{2}y)}^2=(8x{)}^2+{\left(\frac{3}{2}y\right)}^2+2\times 8x\times \frac{3}{2}y=64x^2+\frac{9}{4}{y}^2+24xy=64x^2+24xy+\frac{9}{4}{y}^2$

(vi) $(607{)}^2=(600+7{)}^2=(600{)}^2+(7{)}^2+2\left(600\right)\left(7\right)=360000+49+8400=368449$

(vii) \((391)^2 = (400 - 9)^2 = (400)^2 + 9^2 - 2(400)(9)\\ = 160000 + 81 - 7200 = 152881)

(viii) $(9.7{)}^2=(10-.3{)}^2=(10{)}^2+(.3{)}^2-2\left(10\right)\left(.3\right)=100+.09-6=100.09-6.00=94.09$