>Radical Equations>What Are Radical Equations?

What Are Radical Equations?

Radical equations are equations where there is an $x$ in the radicand of a square root. With this type of equation, we might end up with extraneous solutions, therefore testing answers to the equation is a part of the solution process.

Here is how you do it:

Rule

Radical equations

1.: Isolate the square root on one side.
2.: Square both sides.
3.: Calculate and solve the equation normally.
4.: Test the answer to the equation.

Testing the answer to the equation is a part of the solving method for these equations. It often happens that you get extraneous solutions. It’s in “Testing the answer” where you detect and remove these extraneous solutions.

Example 1

Solve the equation $x - \sqrt{x^{2} - 1} = 1$

\begin{array}{l} x - \sqrt{x^{2} - 1} & = 1 \\ - \sqrt{x^{2} - 1} & = 1 - x \\ x - 1 & = \sqrt{x^{2} - 1} \\ {(x - 1)}^{2} & = {\sqrt{x^{2} - 1}}^{2} \\ x^{2} - 2 x + 1 & = x^{2} - 1 \\ - 2 x & = - 2 \\ x & = 1 \end{array}

Now you test the answer:

\begin{array}{l} Left side & = 1 - \sqrt{1^{2} - 1} = 1 \\ Right side & = 1 \end{array}

Since Left side $=$ Right side, the answer is $x = 1$ .

Example 2

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

\begin{array}{l} \sqrt{x + 1} & = x - 3 \\ {\sqrt{x + 1}}^{2} & = {(x - 3)}^{2} \\ x + 1 & = x^{2} - 6 x + 9 \\ x^{2} - 7 x + 8 & = 0 \end{array}

Now you solve the quadratic expression with the quadratic formula, or with inspection. We’ll use the quadratic formula:

\begin{array}{l} x & = \frac{7 \pm \sqrt{{(7)}^{2} - 4 \cdot 1 \cdot (8)}}{2 \cdot 1} \\ = \frac{7 \pm \sqrt{49 - 32}}{2} \\ = \frac{7 \pm \sqrt{17}}{2}, \end{array}

Possible solutions are then:

\begin{array}{l}  \end{array}

x = 7 + 17 2 ≈ 1.438 and x = 7 −17 2 ≈ 5.562

Now you have to test the answer by inserting

x = \frac{7 + \sqrt{17}}{2} \approx 1.438

(If you get a small deviation on the decimals when you test the solutions, it is due to the rounding):

\begin{array}{l} Left side & = \sqrt{1.438 + 1} = 1.561 \\ Right side & = 1.438 - 3 = - 1.562 \end{array}

Since Left side $\neq$ Right side is $x = \frac{7 + \sqrt{17}}{2} \approx 1.438$ an extraneous solution.

Now you insert

x = \frac{7 - \sqrt{17}}{2} \approx 5.562

(If you get a small deviation on the decimals when you test the solutions, it is the rounding).

\begin{array}{l} Left side = \sqrt{5.562 + 1} = 2.562 \\ Right side = 5.562 - 3 = 2.562 \end{array}

Since Left side $=$ Right side, then $x = \frac{7 - \sqrt{17}}{2} \approx 5.562$ is a solution.

The solution to the equation is then

x = \frac{7 - \sqrt{17}}{2} \approx 5.562 .

Note! Testing the answer is a part of the solving method, because squaring can sometimes lead to extraneous solutions.

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