Question

Find the square of :
(i) $x+3y$
(ii) $2x-5y$
(iii) $a+\frac{1}{5a}$
(iv) $2a-1a$

Answer 1

Since we know that

$(a+b{)}^2=a^2+2ab+b^2$

$(a-b{)}^2=a^2-2ab+b^2$

Now,

$\left(i\right)$

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

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

$\left(ii\right)$

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

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

$\left(iii\right)$

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

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

$\left(iv\right)$

$(2a-1a)=(1a{)}^2$

$(1a{)}^2=1a^2$

Answer