# Question 07e50

Sep 29, 2017

$x = 3$. Don't be fooled by a phantom(see below).

#### Explanation:

You have to square the formula to remove sqrt sign, but if
you calculate ${\left(2 x + \sqrt{x + 1}\right)}^{2}$ will yield $4 {x}^{2} + 4 x \sqrt{x + 1} + x + 1$ and make matters worse.

First, move the term $2 x$ to the right side:
$\sqrt{x + 1} = 8 - 2 x$

Then square the formula:
${\left(\sqrt{x + 1}\right)}^{2} = {\left(8 - 2 x\right)}^{2}$
$x + 1 = 64 - 32 x + 4 {x}^{2}$

And reduce this:
$4 {x}^{2} - 33 x + 63 = 0$

Use the quadratic formula and you will reach
x=(33±sqrt(33^2-4xx4xx63))/8
 x=(33±9)/8#
$x = \frac{21}{4} , 3$
(or you can factor the equation to $\left(4 x - 21\right) \left(x - 3\right) = 0$)
Important: This is not the final answer and you need to check it!

As the range of $\sqrt{x + 1}$ is $\sqrt{x + 1} \ge 0$ , the condition
$8 - 2 x > 0$ must be satisfied. $x = 3$ meets the
condition but $x = \frac{21}{4} = 5.25$ does not.

Therefore the root of this equation is $x = 3$.