# How do you solve 5+(sqrt(x+2))=8+(sqrt(x-7))?

Jul 6, 2016

$x = 7.$

#### Explanation:

$5 + \sqrt{x + 2} = 8 + \sqrt{x - 7}$
$\therefore \sqrt{x + 2} - \sqrt{x - 7} = 3$
$\therefore {\left\{\sqrt{x + 2} - \sqrt{x - 7}\right\}}^{2} = {3}^{2}$
$\therefore x + 2 - 2 \cdot \sqrt{x + 2} \cdot \sqrt{x - 7} + x - 7 = 9$
$\therefore 2 x - 14 = 2 \cdot \sqrt{x + 2} \cdot \sqrt{x - 7}$
$\therefore x - 7 = \sqrt{x + 2} \cdot \sqrt{x - 7} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(I\right)$
$\therefore {\left(\sqrt{x - 7}\right)}^{2} = \sqrt{x + 2} \cdot \sqrt{x - 7} \ldots \ldots \ldots \ldots \ldots . . \left(I I\right)$

And, here is a Word of Caution : We can not cancel out (x-7) directly from both sides without checking whether (x-7)=0.

The reason behind this is, in case, $\left(x - 7\right) = 0 ,$ cancelling it directly from both sides means that we are dividing both sides by 0, which is not possible!

Accordingly, we have to consider two cases :$\left(1\right) : \left(x - 7\right) = 0$, & $: \left(2\right) : \left(x - 7\right) \ne 0.$

$C a s e : \left(1\right) : \left(x - 7\right) = 0.$

In this case, we see that $\left(I I\right)$ is satisfied, and hence, $x = 7$ is a soln.

$C a s e : \left(2\right) : \left(x - 7\right) \ne 0.$

In this case, now we can divide both sides by $\left(x - 7\right) ,$ {bcz., it is not zero] and get,

$\sqrt{x - 7} = \sqrt{x + 2} ,$ squaring which, we have an impossibility $- 7 = 2.$

Hence, $x = 7$ is the only soln.

Another way to solve it, is, to square $\left(I\right)$, to get,
${\left(x - 7\right)}^{2} - \left(x + 2\right) \left(x - 7\right) = 0$
$\therefore \left(x - 7\right) \left\{\left(x - 7\right) - \left(x + 2\right)\right\} = 0$
$\therefore \left(x - 7\right) \left(- 9\right) = 0$
$\therefore x = 7.$

How's that? Enjoyable Maths.!