Question

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

• A. $={\frac{1}{16}a}^2+{\frac{1}{4}b}^2+1-\frac{1}{4}ab-b+\frac{1}{2}a$
• B. $={\frac{1}{8}a}^2+{\frac{1}{2}b}^2+1-\frac{1}{8}ab-b+\frac{1}{4}a$
• C. $={\frac{1}{16}a}^2+{\frac{1}{4}b}^2+1-\frac{1}{4}ab-b+\frac{1}{16}a$
• D. $={\frac{1}{2}a}^2+{\frac{1}{8}b}^2+1-\frac{1}{8}ab-b+\frac{1}{16}a$

Answer 1

The correct option is A

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

${(\frac{1}{4}a-\frac{1}{2}b+1)}^2$

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

$a=\frac{1}{4}a,b=-\frac{1}{2}b,c=1$

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

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