# How do you solve sqrt(3x-2)=1+sqrt(2x-3)?

Sep 29, 2016

$x$ = 2 and 6

#### Explanation:

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

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Square both sides

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

The $- 2$ on both sides cancel out

$3 x \cancel{- 2} \text{ "=" } 2 x \cancel{- 2} + 2 \sqrt{2 x - 3}$

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

Subtract $2 x$ from both sides

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

Divide both sides by 2

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

Square both sides

${x}^{2} / 4 = 2 x - 3$

giving:$\text{ } {x}^{2} / 4 - 2 x + 3 = 0$

Using the standardised formula $y = a {x}^{2} + b x + c$

$a = \frac{1}{4} \text{; "b=-2"; } c = 3$

where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\implies x = \frac{+ 2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(\frac{1}{4}\right) \left(3\right)}}{2 \left(\frac{1}{4}\right)}$

$\implies x = 4 \pm \frac{\sqrt{4 - \frac{12}{4}}}{\frac{1}{2}}$

$x = 4 \pm 2$

color(brown)(x=2" and "x=6

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{However:}}$

Consider $\sqrt{2 x - 3}$

For $x \in \mathbb{R} : 2 x \ge 3$ otherwise $2 x - 3$ is negative

$\textcolor{b r o w n}{\implies x \ge \frac{3}{2}}$

,..............................................................................................
Consider $\sqrt{3 x - 2}$

For $x \in \mathbb{R} : 3 x \ge 2$ otherwise $3 x - 2$ is negative

$\textcolor{b r o w n}{\implies x \ge \frac{2}{3}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

$\textcolor{b r o w n}{\text{If "x<3/2" then at least one of the roots is negative}}$

$\textcolor{b r o w n}{\text{As both "x=2" and "x=6" are greater than } \frac{3}{2}}$$\textcolor{b r o w n}{\text{both are feasible solutions}}$

" "color(green)(bar(ul(|color(white)(2/2)x=2" and "x=6" "|))

Sep 29, 2016

$\textcolor{g r e e n}{x = 2 \textcolor{b l a c k \text{ or }}{x} = 6}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \sqrt{3 x - 2} = 1 + \sqrt{2 x - 3}$

Square both sides
$\textcolor{w h i t e}{\text{XXX}} 3 x - 2 = 1 + 2 \sqrt{2 x - 3} + 2 x - 3$

Simplify
$\textcolor{w h i t e}{\text{XXX}} 3 x - 2 = 2 \sqrt{2 x - 3} + 2 x - 2$

Isolate the term with the root on one side
$\textcolor{w h i t e}{\text{XXX}} x = 2 \sqrt{2 x - 3}$

Square both sides
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 8 x - 12$

Convert to standard form
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 8 x + 12 = 0$

Factor to get
$\textcolor{w h i t e}{\text{XXX}} \left(x - 2\right) \left(x - 6\right) = 0$

$\Rightarrow x = 2 \mathmr{and} x = 6$