Question

Part one is -3 and 7
part two is -3

Answer 1

Given equation:

`√(30-2x) =x-3`

Square both sides:

`\implies (√(30-2x))^2 =(x-3)^2`

`\implies 30-2x =(x-3)^2`

Expand the right side:

`\implies 30-2x =x^2-6x+9`

Add
`2x` to both sides:

`\implies 30 =x^2-4x+9`

Subtract 30 from both sides:

`\implies x^2-4x-21=0`

Factor:

`\implies (x-7)(x+3)=0`

Therefore,

`x-7=0 \implies x =7`

`x+3=0 \implies x=-3`

Substitute both values of
`x` into the original equation to check:

`x=7 :`

`\implies√(30-2(7)) =7-3`

`\implies 4=4`

correct!

`x=-3`

`\implies√(30-2(-3)) =-3-3`

`\implies 6 \\eq =-6`

incorrect!

Therefore,
`x=7` is the only correct solution, and
`x=-3` is the extraneous solution

**Extraneous solution: a root of a transformed equation that is not a root of the original equation**