Question

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

Answer 1

`x=2/3`

Explanation:

Given:

`sqrt(x^2+5) = 3-x`

Square both sides (noting that this may introduce spurious solutions) to get:

`x^2+5 = 9-6x+x^2`

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

`5 = 9-6x`

Subtract `9` from both sides to get:

`-4 = -6x`

Divide both sides by `-6` to get:

`2/3 = x`

This is a valid solution of the original equation:

`sqrt((2/3)^2+5) = sqrt(4/9+5) = sqrt(49/9) = 7/3 = (9-2)/3 = 3-2/3`

Answer 2

`x=2/3`

Explanation:

`color(blue)(sqrt(x^2+5)=3-x`

`"Square both sides to cancel out the radical sign-"` `sqrt`

`rarr(sqrt(x^2+5))^2=(3-x)^2`

`rarrx^2+5=(3-x)^2`

`"Use the formula "` `color(orange)((a-b)^2=a^2-2ab+b^2`

`rarrx^2+5=3^2-2(3)(x)+x^2`

`rarrx^2+5=9-6x+x^2`

`"Cancel"` `x^2` `"both sides"`

`rarrcancel(x^2)+5=9-6x+cancel(x^2`

`rarr5=9-6x`

`"Rewrite the equation"`

`rarr6x=9-5`

`rarr6x=4`

`color(green)(rArrx=4/6=2/3`