Question

How do you solve the equation and identify any extraneous solutions for `sqrt(x^2+5)=3-x`?

Answer 1

We can square both sides of the equation. But we have to remember that the outcome of a radical is allways `>=0`

Explanation:

This means that `3-x>=0->x<=3`

Now for the squaring (which annulls the root at the left):
`x^2+5=(3-x)^2->`
`x^2+5=x^2-6x+9->`
`5=-6x+9->6x=4->x=2/3`

And this fits with the condition that `x<=3`
(it also checks if you put the number back into the equation)