Question

${\left(\frac{a}{4}-\frac{b}{2}+1\right)}^2$expand it with suitable identities .

Answer 1

Solve the given identity : Here we use the identity $({\text{x+y+z)}}^2{\text{=x}}^2{\text{+y}}^2{\text{+z}}^2\text{+2xy+2yz+2zx}$

Now we have $\text{x=}\frac{\text{a}}{4}\text{,y=}\frac{\text{-b}}{2}\text{,z=1}$

So ${(\frac{\text{a}}{4}-\frac{\text{b}}{2}+1)}^2=(\frac{\text{a}}{4}{)}^2+{\left(\frac{-\text{b}}{2}\right)}^2+1^2+2\times \frac{\text{a}}{4}\times (-\frac{\text{b}}{2})+2(-\frac{\text{b}}{2})\times 1+2\times 1\times (\frac{\text{a}}{4})$

$=\frac{{\text{a}}^2}{16}+\frac{{\text{b}}^2}{4}+1-\frac{\text{ab}}{4}-\text{b+}\frac{\text{a}}{2}$

Hence ${(\frac{\text{a}}{4}-\frac{\text{b}}{2}+1)}^2$$=\frac{{\text{a}}^2}{16}+\frac{{\text{b}}^2}{4}+1-\frac{\text{ab}}{4}-\text{b+}\frac{\text{a}}{2}$