Start by squaring both sides to eliminate some of the square roots!

3x + 4 + x + 2 - 2√(3x + 4)(x + 2) = 4

Tidy up and square again...

4x + 2 = 2√(3x + 4)(x + 2)

2x + 1 = √((3x + 4)(x + 2))

(2x + 1)² = (3x + 4)(x + 2)

4x² + 4x + 1 = 3x² + 10x + 8

Now collect terms to form a standard quadratic, and solve.

x² - 6x – 7 = 0

(x – 7)(x + 1) = 0

So it looks as if x = 7 and x = -1 form the solution to the original equation

But hold on!

Check by substituting these values back into the given equation: x = - 1 is fine, but if you substitute x = 7 you get

5 + 3 which is definitely not 2.

What’s happened?

We should look at the line 2x + 1 = √((3x + 4)(x + 2)).

The notation √ always means the positive square root, so as well as needing (3x + 4)(x + 2) to be positive so it can be square-rooted, we need 2x + 1 to be positive as well – but if x = -1 this doesn’t happen.

Try solving the equation √(3x + 4) + √(x + 2) = 2 and compare the steps of your working. Where do the steps of the two problems become the same and what operation has made this happen?

Moral: be careful of the signs when you square expressions....

Challenge 287 : The root of the problem...