Question

How do you evaluate `(2g - 1) ^2 - 2g + g ^2` for g = 3?

Answer 1

For `g=3`, we have `(2g-1)^2-2g+g^2=28`

Explanation:

Putting `g=3` in `(2g-1)^2-2g+g^2`,

we get `(2xx3-1)^2-2xx3+3^2`

= `(6-1)^2-6+9`

= `5^2+3`

= `25+3`

= `28`