Question

Expand the following, using suitable identities: $\left[\right(\frac{1}{4})a-{\left(\frac{1}{2}\right)b+1)}^2$

Answer 1

Given : $\left[\right(\frac{1}{4})a-{\left(\frac{1}{2}\right)b+1)}^2$

$$\begin{gathered} U\sin gidentity{(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx \\ Here,x=\left(\frac{1}{4}\right)a,y=(-\frac{1}{2})b,z=1 \end{gathered}$$

$$\begin{gathered} \left[\right(\frac{1}{4})a-{\left(\frac{1}{2}\right)b+1)}^2 \\ ={\left(\frac{1}{4}a\right)}^2+{\left(-\frac{1}{2}b\right)}^2+1^2+\left(2\times \frac{1}{4}a\times -\frac{1}{2}b\right)+\left(2\times -\frac{1}{2}b\times 1\right)+\left(2\times 1\times \frac{1}{4}a\right) \end{gathered}$$

$$\begin{gathered} =\frac{1}{16}{a}^2+\frac{1}{4}{b}^2+1^2-\frac{2}{8}ab-\frac{2}{2}b+\frac{2}{4}a \\ =\frac{1}{16}{a}^2+\frac{1}{4}{b}^2+1^{}-\frac{1}{4}ab-b+\frac{1}{2}a \end{gathered}$$