How do you solve sqrt(4x+17) +sqrt(x +7) =sqrt(x + 2)?

Aug 11, 2015

$x = \emptyset$

Explanation:

Start by writing down the conditions that $x$ must meet in order to be considered a valid solution

• $4 x + 17 \ge 0 \implies x \ge - \frac{17}{4}$
• $x + 7 \ge 0 \implies x \ge - 7$
• $x + 2 \ge 0 \implies x \ge - 2$

These conditions are determined by the fact that the expressions under the radicals must be positive if you're dealing with real numbers.

Combine all three conditions to get $x \ge - 2$.

Another important thing to notice here is that for any $x$ that qualifies as a valid solution, you know that

$4 x + 17 > x + 7$

Moreover,

$x + 7 > x + 2$

This actually tells you that the equation has no valid solutions among real numbers, since two positive numbers cannot be added to give a smaller positive number.

In other simply, any solution that you will come about by solving this equation will be extraneous.

Now, square both sides of the equation to reduce the number of radical terms from three to one.

${\left(\sqrt{4 x + 17} + \sqrt{x + 7}\right)}^{2} = {\left(\sqrt{x + 2}\right)}^{2}$

${\left(\sqrt{4 x + 17}\right)}^{2} + 2 \sqrt{\left(4 x + 17\right) \left(x + 7\right)} + {\left(\sqrt{x + 7}\right)}^{2} = x + 2$

$4 x + 17 + 2 \sqrt{\left(4 x + 17\right) \left(x + 7\right)} + \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 7 = \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 2$

This is equivalent to

$2 \sqrt{\left(4 x + 17\right) \left(x + 7\right)} = - 4 x - 22$

Square both sides of the equation again to get rid of the last radical term

(2sqrt((4x+17)(x+7)))^2 = (-4x - 22)""^2

$4 \left(4 x + 17\right) \left(x + 7\right) = 16 {x}^{2} + 176 x + 484$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{16 {x}^{2}}}} + 180 x + 476 = \textcolor{red}{\cancel{\textcolor{b l a c k}{16 {x}^{2}}}} + 176 x + 484$

This is equivalent to

$4 x = 8 \implies x = \frac{8}{4} = \textcolor{g r e e n}{2}$

Notice that $x = 2$ satisfies the condition $x \ge - 2$, which means that it qualifies as a valid solution. However, a quick check confirms that it is not a valid solution

$\sqrt{4 \cdot \left(2\right) + 17} + \sqrt{2 + 7} = \sqrt{2 + 2}$

$\sqrt{25} + \sqrt{9} = \sqrt{4}$

$5 + 3 \textcolor{R e d}{\ne} 2 \to x = 2$ is an extraneous solution.