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Exercises Equations and inequalities

Example 6 from Algebra and Trigonometry, 2.6 Other Types of Equations

Solving an Equation with One Radical

Solve $\sqrt{15 - 2 x} = x .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
The radical is already isolated on the left side of the equal side, so proceed to square both sides.

$\begin{matrix} \sqrt{15 - 2 x} & = & x \\ {(\sqrt{15 - 2 x})}^{2} & = & {(x)}^{2} \\ 15 - 2 x & = & x^{2} \end{matrix}$

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

$\begin{matrix} 0 & = & x^{2} + 2 x - 15 \\ = & (x + 5) (x - 3) \\ x & = & −5 \\ x & = & 3 \end{matrix}$

The proposed solutions are $−5$ and $3.$ Let us check each solution back in the original equation. First, check $x = −5.$
$\begin{matrix} \sqrt{15 - 2 x} & = & x \\ \sqrt{15 - 2 (- 5)} & = & −5 \\ \sqrt{25} & = & −5 \\ 5 & \neq & -5 \end{matrix}$

This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.

Check $x = 3.$

$\begin{matrix} \sqrt{15 - 2 x} & = & x \\ \sqrt{15 - 2 (3)} & = & 3 \\ \sqrt{9} & = & 3 \\ 3 & = & 3 \end{matrix}$
The solution is $3.$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the radical equation: $\sqrt{x + 3} = 3 x - 1$

How did it go?

PreviousSolve a Polynomial by Grouping Next Solving a Radical Equation Containing Two Radicals

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 2.6, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.