Solving Radical Equations and Identifying Extraneous Solutions

This algebra lesson is a sample from The Ultimate Crash Course for STEM Majors. The example demonstrates how to solve an equation containing two radicals, why squaring both sides can be useful, and why students must be aware of extraneous solutions. Explore the complete mathematics, physics, and STEM lesson collection through The Ultimate Crash Course for STEM Majors .

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Radical Equations and Extraneous Solutions

\[ \sqrt{2x+3}-\sqrt{x+1}=1 \]

Before solving anything, take note of extraneous solutions. These are solutions that will not satisfy the original equation. Usually, the above equation would be presented as something of the nature of letting \(f(x)=\sqrt{2x+3}-\sqrt{x+1}\), \(g(x)=1\). Solve \(f(x)=g(x)\).

When looking at equations with radicals, you may have to use your brain!

We always need to know the wrong roads to take so to avoid them on exams.

Consider the form before jumping in. In this case, we have \(a-b\), and we know from reading the book that \((a-b)(a+b)=a^2-b^2\). [Conjugate] But this doesn’t help.

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

\[ (2x+3)-(x+1)=\sqrt{2x+3}+\sqrt{x+1} \]

All this did was flip flop the equation.

Squaring Both Sides of a Radical Equation

In general, when there are two radicals, first square each side to bring it to one radical

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

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

Using \((a-b)^2=a^2-2ab+b^2\), \(\sqrt{a}\sqrt{b}=\sqrt{ab}\) [\(\Rightarrow\) implies \(p=q\Rightarrow q=p\)]

\[ \Rightarrow (2x+3)-2\sqrt{(x+1)(2x+3)}+(x+1)=1 \]

\[ \Rightarrow -2\sqrt{(x+1)(2x+3)}+3x+4=1 \Rightarrow -2\sqrt{(x+1)(2x+3)}=-3-3x \]

\[ \Rightarrow \sqrt{(x+1)(2x+3)} = -\frac{1}{2}(-3-3x) = \frac{(-3)}{2}(1+x)x = \frac{3}{2}(1+x) \]

\[ \Rightarrow \sqrt{(x+1)(2x+3)} = \frac{3}{2}(1+x) \]

Isolating the Radical and Squaring Again

Now that we have isolated the single radical, we can square each side again.

\[ \Rightarrow \left(\sqrt{(x+1)(2x+3)}\right)^2 = \left[\frac{3}{2}(1+x)\right]^2 \]

\[ \Rightarrow (x+1)(2x+3) = \left(\frac{3}{2}\right)^2(1+x)^2 \Rightarrow (x+1)(2x+3) = \frac{9}{4}(1+x)^2 \]

\[ \Rightarrow 4(x+1)(2x+3) = 9(1+x)(1+x) \Rightarrow 4[2x^2+3x+2x+3] = 9[1+2x+x^2] \]

Learn More Algebra, Mathematics, Physics, and Engineering

This lesson on solving radical equations and recognizing extraneous solutions is a sample from The Ultimate Crash Course for STEM Majors. The series contains worked mathematics, physics, engineering, and STEM lessons designed to demonstrate the mathematical process step by step.