Question

How do you solve `(w - 7) ^ { 2} = 2w ^ { 2} - 20w + 57`?

Answer 1

`w =2 or w =4`

Explanation:

You are dealing with a quadratic equation so to solve it you need to set it equal to `0`.

Multiply out the bracket:

`color(blue)((w-7)^2) = 2w^2 -20w+57`

`color(blue)((w-7)(w-7)) = 2w^2 -20w+57`

`color(blue)(w^2-14w+49) =2w^2-20w+57`

`color(white)(xxxxxxxxx)0 = 2w^2-20w+57 color(blue)(-w^2+14w-49)`

`color(white)(xxxxxxxxx)0 = w^2-6w+8`

Find factors of `8` which add up to `6`

`(w-4)(w-2)=0`

Set each factor equal to `0`

If `w -4 =0 " "rArr w =4`

If `w -2 =0 " "rArr w =2`