Example 7 from Algebra and Trigonometry, 2.6 Other Types of Equations

Solving a Radical Equation Containing Two Radicals

Solve $\sqrt{2 x + 3} + \sqrt{x - 2} = 4.$

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
As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.

$\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} & = & 4 \\ \sqrt{2 x + 3} & = & 4 - \sqrt{x - 2} & Subtract \sqrt{x - 2} from both sides . \\ {(\sqrt{2 x + 3})}^{2} & = & {(4 - \sqrt{x - 2})}^{2} & Square both sides . \end{matrix}$
Use the perfect square formula to expand the right side: ${(a - b)}^{2} = a^{2} −2 a b + b^{2} .$

$\begin{matrix} 2 x + 3 & = & {(4)}^{2} - 2 (4) \sqrt{x - 2} + {(\sqrt{x - 2})}^{2} \\ 2 x + 3 & = & 16 - 8 \sqrt{x - 2} + (x - 2) \\ 2 x + 3 & = & 14 + x - 8 \sqrt{x - 2} & Combine like terms . \\ x - 11 & = & −8 \sqrt{x - 2} & Isolate the second radical . \\ {(x - 11)}^{2} & = & {(−8 \sqrt{x - 2})}^{2} & Square both sides . \\ x^{2} - 22 x + 121 & = & 64 (x - 2) \end{matrix}$
Now that both radicals have been eliminated, set the quadratic equal to zero and solve.

$\begin{matrix} x^{2} - 22 x + 121 & = & 64 x - 128 \\ x^{2} - 86 x + 249 & = & 0 \\ (x - 3) (x - 83) & = & 0 & Factor and solve . \\ x & = & 3 \\ x & = & 83 \end{matrix}$

The proposed solutions are $3$ and $83.$ Check each solution in the original equation.
$\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} & = & 4 \\ \sqrt{2 x + 3} & = & 4 - \sqrt{x - 2} \\ \sqrt{2 (3) + 3} & = & 4 - \sqrt{(3) - 2} \\ \sqrt{9} & = & 4 - \sqrt{1} \\ 3 & = & 3 \end{matrix}$
One solution is $3.$
Check $x = 83.$

$\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} & = & 4 \\ \sqrt{2 x + 3} & = & 4 - \sqrt{x - 2} \\ \sqrt{2 (83) + 3} & = & 4 - \sqrt{(83 - 2)} \\ \sqrt{169} & = & 4 - \sqrt{81} \\ 13 & \neq & - 5 \end{matrix}$
The only solution is $3.$ We see that $83$ is an extraneous solution.

Your turn

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

Solve the equation with two radicals: $\sqrt{3 x + 7} + \sqrt{x + 2} = 1.$

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Solution

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Solution

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