How do you solve  sqrt (9x+1) + 6= 2x?

Jul 24, 2015

Rewrite this equation as a quadratic and proceed to solve for $x$.

Explanation:

Before doing anything else, isolate the radical term on one side of the equation.

To do that, add $- 6$ to both sides of the equation

$\sqrt{9 x + 1} + \cancel{6} - \cancel{6} = 2 x - 6$

$\sqrt{9 x + 1} = 2 \left(x - 3\right)$

Now square both sides of the equation to get rid of the radical term

${\left(\sqrt{9 x + 1}\right)}^{2} = {\left[2 \left(x - 3\right)\right]}^{2}$

$9 x + 1 = {2}^{2} \cdot {\left(x - 3\right)}^{2}$

$9 x + 1 = 4 \cdot \left({x}^{2} - 6 x + 9\right)$

$9 x + 1 = 4 {x}^{2} - 24 x + 36$

Move all your terms to one side of the equation to get

$4 {x}^{2} - 33 x + 35 = 0$

Use the quadratic formula to determine the two solutions ofr this equation

${x}_{1 , 2} = \frac{33 \pm \sqrt{1089 - 4 \cdot 4 \cdot 35}}{8}$

${x}_{1 , 2} = \frac{33 \pm \sqrt{529}}{8} = \frac{33 \pm 23}{8} \implies \left\{\begin{matrix}{x}_{1} = \frac{33 + 23}{8} = \textcolor{b l u e}{7} \\ {x}_{2} = \frac{33 - 23}{8} = \textcolor{\mathmr{and} a n \ge}{\frac{5}{4}}\end{matrix}\right.$

Check to see if both solutions satisfy the original equation. Since both solutions are positive, the radical term will always be positive, so all you really need to check is whether or not the right side of the equation is positive as well.

For ${x}_{1}$ you have

$2 \cdot \textcolor{b l u e}{7} - 6 > 0$

$8 > 0 \to {x}_{1}$ is $\textcolor{g r e e n}{\text{valid}}$

For ${x}_{2}$ you have

$2 \cdot \textcolor{\mathmr{and} a n \ge}{\frac{5}{4}} - 6 > 0$

$- \frac{7}{2} \cancel{>} 0 \to {x}_{2}$ is $\textcolor{red}{\text{not valid}}$