Question

Find the square of :

(i) $x+3y$

(ii) $2x-5y$

(iii) $a+\frac{1}{5a}$

(iv) $2a-\frac{1}{a}$

(v) $x-2y+1$

(vi) $3a-2b-5c$

(vii) $2x+\frac{1}{x}+1$

(viii) $5-x+\frac{2}{x}$

(ix) $2x-3y+z$

(x) $x+\frac{1}{x}-1$

Answer 1

(i) $(x+3y{)}^2=(x{)}^2+(3y{)}^2+2\times x\times 3y$

$=x^2+9y^2+6xy$.

(ii) $(2x-5y{)}^2=(2x{)}^2+(5y{)}^2-2\times 2x\times 5y$

$=4x^2+25y^2-20xy$.

(iii) ${(a+\frac{1}{5a})}^2=(a{)}^2+{\left(\frac{1}{5a}\right)}^2+2\times a\times \frac{1}{5a}$

$=a^2+\frac{1}{25a^2}+\frac{2}{5}$.

(iv) ${(2a-\frac{1}{a})}^2=(2a{)}^2+{\left(\frac{1}{a}\right)}^2-2\times 2a\times \frac{1}{a}$

$=4a^2+\frac{1}{a^2}-4$.

(v) $(x-2y+1{)}^2=(x{)}^2+(-2y{)}^2+(1{)}^2+2\times x\times -2y+2\times (-2y)\times 1+2\times 1\times x$

$=x^2+4y^2+1-4xy-4y+2x$.

(vi) $(3a-2b-5c{)}^2=(3a{)}^2+(-2b{)}^2+(-5c{)}^2+2\times 3a\times -2b+2\times (-2b)(-5c)+2\times -5c\times 3a$

$=9a^2+4b^2+25c^2-12ab+20bc-30ca$.

(vii) $(2x+\frac{1}{x}+1)=(2x{)}^2+{\left(\frac{1}{x}\right)}^2+(1{)}^2+2\times 2x\times \frac{1}{x}+2\times \frac{1}{x}\times 1+2\times 1\times 2x$

$=4x^2+\frac{1}{x^2}+1+4+\frac{2}{x}+4x$

$=4x^2+\frac{1}{x^2}+5+\frac{2}{x}+4x$.

(viii) ${(5-x+\frac{2}{x})}^2=(5{)}^2+(-x{)}^2+{\left(\frac{2}{x}\right)}^2+2\times 5+\times (-x)+2(-x)\times \frac{2}{x}+2\times \frac{2}{x}\times 5$

$=25+x^2+\frac{4}{x^2}-10x-4+\frac{20}{x}$

$=21+x^2+\frac{4}{x^2-10x+\frac{20}{x}}$.

(ix) $(2x-3y+z{)}^2=(2x{)}^2+(-3y{)}^2+(z{)}^2+2\times 2x\times -3y+2(-3y)\times z+2\times z\times 2x$

$=4x^2+9y^2+z^2-12xy-6yz+4zx$.

(x) ${(x+\frac{1}{x}-1)}^2=(x{)}^2+{\left(\frac{1}{x}\right)}^2+(-1{)}^2+2\times x\times \frac{1}{x}+2\times \frac{1}{x}\times (-1)+2(-1)\times x$

$=x^2+\frac{1}{x^2}+1+2-\frac{2}{x}-2x$

$=x^2+\frac{1}{x^2+3-\frac{2}{x}-2x}$.