How do you solve 3sqrtk+1=6 and find any extraneous solutions?

Aug 26, 2016

$\frac{25}{9}$

Explanation:

We require to isolate the square root term.

subtract 1 from both sides

$\Rightarrow 3 \sqrt{k} + \cancel{1} - \cancel{1} = 6 - 1 \Rightarrow 3 \sqrt{k} = 5$

divide both sides by 3

$\Rightarrow \frac{{\cancel{3}}^{1} \sqrt{k}}{\cancel{3}} ^ 1 = \frac{5}{3} \Rightarrow \sqrt{k} = \frac{5}{3}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sqrt{k} \times \sqrt{k} = {\left(\sqrt{k}\right)}^{2} = k} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Squaring both sides to obtain k

$\Rightarrow {\left(\sqrt{k}\right)}^{2} = {\left(\frac{5}{3}\right)}^{2}$

$\Rightarrow k = {5}^{2} / {3}^{2} = \frac{25}{9}$

There are no extraneous solutions.