Question

How do you solve `\sqrt { 14x + 8} = x + 4`?

Answer 1

`x = 2" "` or `" "x = 4`

Explanation:

Given:

`sqrt(14x+8) = x+4`

Square both sides of the equation - noting that this may introduce extraneous solutions - to get:

`14x+8 = x^2+8x+16`

Subtract `14x+8` from both sides to get:

`0 = x^2-6x+8 = (x-2)(x-4)`

So:

`x = 2" "` or `" "x = 4`

These are both valid solutions of the original equation, since they both result in positive values of `x+4`, while extraneous solutions would be related to the negative square root of `14x+8`.

Want to be sure?

`sqrt(14(color(blue)(2))+8) = sqrt(28+8) = sqrt(36) = 6 = color(blue)(2)+4`

`sqrt(14(color(blue)(4))+8) = sqrt(56+8) = sqrt(64) = 8 = color(blue)(4)+4`

Footnote concerning extraneous solutions

If you are given an equation to solve, then you may attempt to simplify the problem by doing the same thing to both sides of the equation to give you a simpler equation.

If that simplification is reversible (such as adding the same expression to both sides or multiplying both sides by a non-zero constant) then any solution of the resulting equation is a solution of the original one (and vice versa).

If the simplification is not reversible (such as squaring both sides) then solutions of the resulting equation may or may not be solutions of the original equation.

Solutions of the simplified equation that are not solutions of the original are commonly called extraneous solutions or extraneous roots.

To identify which roots are extraneous or not, you can try the solutions in the original equation, or you may find it helpful to identify the specific breakages that may occur. For example, if you have multiplied both sides of a rational equation by `(x-1)` to get a polynomial equation, then `x=1` would be an extraneous solution.

In our example above, an equation formed by squaring both sides of an equation could result in extraneous solutions if the signs of the left and right hand sides were opposite.