# How do you solve sqrt(x+15) - sqrt(2x+7)=1?

Aug 11, 2015

$x = 1$

#### Explanation:

Even before doing any calculations, you know that any possible solution to this equation must satisfy two conditions

• $x + 15 \ge 0 \implies x \ge - 15$
• $2 x + 7 \ge 0 \implies x \ge - \frac{7}{2}$
• $\sqrt{x + 15} > \sqrt{2 x + 7}$

The third condition implies that $x + 15 > 2 x + 7 \implies x < 8$.

Combining these conditions will result in $x \in \left[- \frac{7}{2} , 8\right)$.

These condtions are required because you can't take the square root of a negative number if you're working with real numbers.

In other words, you can only take the square root of positive real numbers; likewise, taking the square root of a positive number will always result in a positive number.

Start by squaring both sides of the equation; this will reduce the number of radical terms from two to one

${\left(\sqrt{x + 15} - \sqrt{2 x + 7}\right)}^{2} = {1}^{2}$

${\left(\sqrt{x + 15}\right)}^{2} - 2 \sqrt{\left(x + 15\right) \left(2 x + 7\right)} + {\left(\sqrt{2 x + 7}\right)}^{2} = 1$

$x + 15 - 2 \sqrt{\left(x + 15\right) \left(2 x + 7\right)} + 2 x + 7 = 1$

Isolate the remaining radical term on one side of the equation

$- 2 \sqrt{\left(x + 15\right) \left(2 x + 7\right)} = 1 - 3 x - 22$

$2 \sqrt{\left(x + 15\right) \left(2 x + 7\right)} = 3 x + 21$

Once again, square both sides of the equation to get rid of the radical lterm

${\left(2 \sqrt{\left(x + 15\right) \left(2 x + 7\right)}\right)}^{2} = {\left(3 x + 21\right)}^{2}$

$4 \left(x + 15\right) \left(2 x + 7\right) = 9 {x}^{2} + 126 x + 441$

$8 {x}^{2} + 148 x + 420 = 9 {x}^{2} + 126 x + 441$

Rearrange this equation to classic quadratic form

${x}^{2} - 22 x + 21 = 0$

You can use the quadratic equation to find the two roots of this quadratic equation

${x}_{1 , 2} = \frac{- \left(- 22\right) \pm \sqrt{{\left(- 22\right)}^{2} - 4 \cdot 1 \cdot 21}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{22 \pm \sqrt{400}}{2} = \frac{22 \pm 20}{2} = \left\{\begin{matrix}{x}_{1} = \frac{22 + 20}{2} = 21 \\ {x}_{2} = \frac{22 - 20}{2} = 1\end{matrix}\right.$

Notice that ${x}_{1}$ does not satisfy the condition $x \in \left[- \frac{7}{2} , 8\right)$, which means that the only valid solution to the original equation is ${x}_{2} = 1$.

More specifically,

$\sqrt{21 + 15} - \sqrt{2 \cdot 21 + 7} = 1$

$6 - 7 \textcolor{red}{\ne} 1 \to {x}_{1}$ is an extraneous solution

$\sqrt{1 + 15} - \sqrt{2 \cdot 1 + 7} = 1$

$4 - 3 = 1 \textcolor{g r e e n}{\sqrt{}}$