# How do you solve sqrt(x+3)-sqrt(x-1)=1?

Apr 19, 2015

the answer is $x = \frac{13}{4}$

First, let's call $a = \sqrt{x + 3}$ and $b = \sqrt{x - 1}$

Now use the rule $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$
so $4 = {a}^{2} - {b}^{2} = a + b$

So we have $a + b = 4$ and $a - b = 1$

We sum and subtract respectively the two equations and we have

$2 a = 5$
$2 b = 3$

So $\sqrt{x + 3} = \frac{5}{2} \implies x = \pm \frac{25}{4} - 3 \implies x \setminus \in \left\{\frac{13}{4} , - \frac{37}{4}\right\}$
And $\sqrt{x - 1} = \frac{3}{2} \implies x = \pm \frac{9}{4} + 1 \implies x \setminus \in \left\{\frac{13}{4} , - \frac{5}{4}\right\}$

so $x = \frac{13}{4}$

Apr 19, 2015

Firstly know that:

$\sqrt{a} \sqrt{b} = \sqrt{a b}$

And also that:

$\sqrt{q} \sqrt{q} = q$

Knowing this, let's find the value of x...

$\sqrt{x + 3} - \sqrt{x - 1} = 1$

${\left(\sqrt{x + 3} - \sqrt{x - 1}\right)}^{2} = {1}^{2}$

$\left(\sqrt{x + 3} - \sqrt{x - 1}\right) \left(\sqrt{x + 3} - \sqrt{x - 1}\right) = 1$

$x + 3 - \sqrt{x + 3} \sqrt{x - 1} - \sqrt{x + 3} \sqrt{x - 1} + \left(x - 1\right) = 1$

$x + 3 - 2 \sqrt{x + 3} \sqrt{x - 1} + x - 1 = 1$

$2 x + 2 - 2 \sqrt{x + 3} \sqrt{x - 1} = 1$

$2 \left(x + 1 - \sqrt{x + 3} \sqrt{x - 1}\right) = 1$

$x + 1 - \sqrt{x + 3} \sqrt{x - 1} = \frac{1}{2}$

$x + 1 - \frac{1}{2} = \sqrt{\left(x + 3\right) \left(x - 1\right)}$

$x + \frac{1}{2} = \sqrt{\left(x + 3\right) \left(x - 1\right)}$

${\left(x + \frac{1}{2}\right)}^{2} = \left(x + 3\right) \left(x - 1\right)$

${x}^{2} + \frac{1}{2} x + \frac{1}{2} x + \frac{1}{4} = {x}^{2} - x + 3 x - 3$

${x}^{2} + x + \frac{1}{4} = {x}^{2} + 2 x - 3$

${x}^{2} - {x}^{2} + \frac{1}{4} + 3 = 2 x - x$

$\therefore x = 3 + \frac{1}{4}$