Question

How do you solve the equation and identify any extraneous solutions for `sqrt(x^2 + 2) = x + 4`?

Answer 1

I found `x=-7/4`

Explanation:

I would square both sides to get:
`x^2+2=(x+4)^2`
`cancel(x^2)+2=cancel(x^2)+8x+16`
`8x=-14`
`x=-14/8`
`x=-7/4`

Answer 2

Square both sides to get:

`x^2+2 = x^2+8x+16`

Hence `x = -7/4`

Explanation:

Try squaring both sides to get:

`x^2+2 = (x+4)^2 = x^2+8x+16`

Subtract `x^2+16` from both sides to get:

`8x = -14`

Divide both sides by `8` to get:

`x = -14/8 = -7/4`

Check:

`sqrt(x^2+2) = sqrt(49/16+32/16) = sqrt(81/16) = 9/4`

`x+4 = -7/4 + 4 = -7/4 + 16/4 = 9/4`

graph{(y - sqrt(x^2+2))*(y - x - 4) = 0 [-9.42, 10.58, -1.8, 8.2]}