# How do you solve sqrt(x+3)-sqrt x=sqrt(4x-5)?

Mar 5, 2016

$x = \frac{16}{11}$

#### Explanation:

This is a tricky equation, so you have first to determine the dominion of it:

$x + 3 \ge 0 \mathmr{and} x > 0 \mathmr{and} 4 x - 5 \ge 0$

$x \ge - 3 \mathmr{and} x > 0 \mathmr{and} x \ge \frac{5}{4} \implies x \ge \frac{5}{4}$

The standard way to solve this type of equations is to square the parcels, admiting that:

$\textcolor{red}{\mathmr{if} a = b \implies {a}^{2} = {b}^{2}}$

However this brings false solutions, because

$\textcolor{red}{\mathmr{if} a = - b \implies {a}^{2} = {b}^{2}}$

So we have to check the solutions after we obtain the results.

So now let's start:

$\sqrt{x + 3} - \sqrt{x} = \sqrt{4 x - 5}$

${\left(\sqrt{x + 3} - \sqrt{x}\right)}^{2} = {\left(\sqrt{4 x - 5}\right)}^{2}$

$x + 3 - 2 \sqrt{\left(x + 3\right) x} + x = 4 x - 5$

Now, you continue to have a "sqrt" in the equation, so you have to square it again. Rearrange the equation in order to isolate the root:

$2 \sqrt{{x}^{2} + 3 x} = 4 x - 5 - x - 3 - x$

$2 \sqrt{{x}^{2} + 3 x} = 2 x - 8$

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

squaring:

${x}^{2} + 3 x = {x}^{2} - 8 x + 16$

Which gives:

$x = \frac{16}{11}$

First 16/11>5/4?(the dominion determined above)

Put them in the same denominator:

(16/11)xx(4/4)>(5/4)xx(11/11) ?

$\frac{64}{44} > \frac{55}{44} , t r u e$

Now, is the solution true?

$\sqrt{\frac{16}{11} + 3} - \sqrt{\frac{16}{11}} = \sqrt{4 \times \frac{16}{11} - 5}$

$\sqrt{\frac{49}{11}} - \sqrt{\frac{16}{11}} = \sqrt{\frac{9}{11}}$

$\frac{\sqrt{49} - \sqrt{16}}{\sqrt{11}} = \sqrt{\frac{9}{11}}$

$\frac{7 - 4}{\sqrt{11}} = \frac{3}{\sqrt{11}} , t r u e$

Mar 5, 2016

$x = \frac{16}{11}$

#### Explanation:

$1$. When dealing with radicals, try to eliminate them first. Thus, start by squaring both sides of the equation.

$\sqrt{x + 3} - \sqrt{x} = \sqrt{4 x - 5}$

${\left(\sqrt{x + 3} - \sqrt{x}\right)}^{2} = {\left(\sqrt{4 x - 5}\right)}^{2}$

$2$. Simplify.

$\left(\sqrt{x + 3} - \sqrt{x}\right) \left(\sqrt{x + 3} - \sqrt{x}\right) = 4 x - 5$

$x + 3 - \sqrt{x \left(x + 3\right)} - \sqrt{x \left(x + 3\right)} + x = 4 x - 5$

$2 x + 3 - \sqrt{{x}^{2} + 3 x} - \sqrt{{x}^{2} + 3 x} = 4 x - 5$

$- 2 \sqrt{{x}^{2} + 3 x} = 2 x - 8$

$\sqrt{{x}^{2} + 3 x} = - \frac{1}{2} \left(2 x - 8\right)$

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

$3$. Since the left side contains a radical, square the whole equation again.

${\left(\sqrt{{x}^{2} + 3 x}\right)}^{2} = {\left(- x + 4\right)}^{2}$

$4$. Simplify.

$\left(\sqrt{{x}^{2} + 3 x}\right) \left(\sqrt{{x}^{2} + 3 x}\right) = \left(- x + 4\right) \left(- x + 4\right)$

${x}^{2} + 3 x = {x}^{2} - 4 x - 4 x + 16$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} + 3 x = \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} - 8 x + 16$

$3 x = - 8 x + 16$

$5$. Solve for $x$.

$11 x = 16$

$\textcolor{g r e e n}{x = \frac{16}{11}}$

$\therefore$, $x$ is $\frac{16}{11}$.