# How do you solve and check for extraneous solutions in sqrt(3x - 2) + 3 =4x?

Aug 10, 2015

$x = 1$

#### Explanation:

Start by isolating the radical term on one side of the equation. This can be done by adding $- 3$ to both sides

$\sqrt{3 x - 2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} = 4 x - 3$

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

Now, because you can't take the squaree root of a negative number if you're working with real numbers, any solution you will find must satisfy two conditions

sqrt( underbrace((3x-2))_(color(orange)("must be positive"))) = overbrace(4x-3)^(color(red)("must be positive"))

• $3 x - 2 \ge 0 \implies x \ge \frac{2}{3}$
• $4 x - 3 \ge 0 \implies x \ge \frac{3}{4}$

Overall, you need $x \ge \frac{3}{4}$.

Square both sides of the equation to remove the radical term

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

$3 x - 2 = 16 {x}^{2} - 24 x + 9$

Rerarrange to get

$16 {x}^{2} - 27 x + 11 = 0$

Use the quadratic formula to find the two solutions to this quadratic equation

${x}_{1 , 2} = \frac{- \left(- 27\right) \pm \sqrt{{\left(- 27\right)}^{2} - 4 \cdot 16 \cdot 11}}{2 \cdot 16}$

${x}_{1 , 2} = \frac{27 \pm \sqrt{25}}{32}$

${x}_{1 , 2} = \frac{27 \pm 5}{32} = \left\{\begin{matrix}{x}_{1} = \frac{27 + 5}{32} = 1 \\ {x}_{2} = \frac{27 - 5}{32} = \frac{11}{16}\end{matrix}\right.$

Notice that only ${x}_{1}$ satisfies the condition $x \ge \frac{3}{4}$, which means that ${x}_{2} = \frac{11}{16}$ will be an extraneous solution.

This is of course because

$\frac{11}{16} < \frac{12}{16} = \frac{3}{4}$

Therefore, your original equation will only have one solution, $x = \textcolor{g r e e n}{1}$.

Do a quick check to make sure that the calculations are corret

$\sqrt{3 \cdot 1 - 2} = 4 \cdot \left(1\right) - 3$

$\sqrt{1} = 1 \iff 1 = 1 \textcolor{g r e e n}{\sqrt{}}$