Radical Equations

A radical equation contains at least one radical sign that includes a variable. For an example you can consider the following equation:

\(\qquad \qquad \qquad \qquad \sqrt{x+2}=x-3\).

Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.

A radical is an exponential number where the exponent is a fraction. For an example, consider \( \sqrt[n]{x^m} = x ^ { m/n } \). A radical equation contains at least one radical sign that includes a variable.

A square root \(\left( \sqrt{x} \right)\) is simplified, when the radicand (value inside the square root sign, which is \(x\) in this case) has no square factors. This can be done effectively by factorizing the radicand first, to know what terms to pull out.

Here is a problem based on the basics of radicals.

The leading step of solving a radical equation is to eliminate the radical exponent to obtain a polynomial equation and check whether the obtained roots satisfy the original equation to escape from extraneous roots.

Following are the steps to solve a radical equation:

1. Eliminate the radicals or rational exponents and obtain a polynomial equation. The key step is to raise each side of the equation to the same power. If \(a = b,\) then \(a^n = b^n\). [Powers property of equality]

2. Then solve the new equation using standard procedures. Before raising each side of an equation to the same power, you should isolate the radical expression on one side of the equation.

3. However, this is not it. When solving radical equations we must always be cautious of extraneous roots. An extraneous root is a root that satisfies the polynomial obtained by squaring, but does not satisfy the original radical equation.

Question: Why do extraneous roots appear?

Answer: Suppose we are solving the radical equation \(A=\sqrt{B},\) where \(A\) and \(B\) are both polynomials. Squaring both sides gives \(A^2=B,\) solving which would give solutions to both \(A=\sqrt{B}\) and \(A=-\sqrt{B}.\) The roots that satisfy \(A=-\sqrt{B}\) are not what we were looking for. Therefore we must never forget to check if the roots we obtained satisfy the original radical equation.

However, we will most often encounter radical equations that contain square roots. The way to solve such equations is simple. Just put the square root on one side of the equation and the non-radicals on the other. Then square both sides and we get a polynomial we can solve!

For demonstration we will take the following example and solve it using the above stated steps:

Solve the equation \(\sqrt{x+3}=x+1.\)

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Step 1:
Squaring both sides (which will get rid of the square root) gives \(x+3=x^2+2x+1.\)

Step 2:
Rearranging terms, we have \[x^2+x-2=(x+2)(x-1)=0 \Rightarrow x=-2 \text{ or } x=1.\]

Step 3:
Observe that \(x=-2\) is an extraneous root because \[\text{LHS}=\sqrt{(-2)+3}=1 \ne (-2)+1=\text{ RHS}.\] Then since \[\text{LHS}=\sqrt{(1)+3}=2=(1)+1=\text{ RHS},\] the solution to the equation \(\sqrt{x+3}=x+1\) is \(x=1.\) \( _\square \)

Try to solve the following problems using the above steps.

This section is dedicated to enhance the problem solving skills through several examples and problems to try.

Solve the equation \(x-2=\sqrt{2x+31}.\)

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Squaring both sides gives \(x^2-4x+4=2x+31.\) Rearranging terms, we have

\[x^2-6x-27=(x+3)(x-9)=0 \Rightarrow x=-3 \text{ or } x=9.\]

Observe that \(x=-3\) is an extraneous root because

\[\text{LHS}=(-3)-2 = -5 \ne \sqrt{2\cdot (-3)+31}=\text{ RHS}.\]

Then since

\[\text{LHS}=(9)-2=7= \sqrt{2\cdot (9)+31}=\text{ RHS},\]

the solution to the equation \(x-2=\sqrt{2x+31}\) is \(x=9.\) \( _\square \)

Solve the equation \(\sqrt{2x^2+7}+1=2x.\)

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The equation \(\sqrt{2x^2+7}+1=2x\) is equivalent to \(\sqrt{2x^2+7}=2x-1.\) Squaring both sides gives \(2x^2+7=4x^2-4x+1.\) Rearranging terms, we have

\[2x^2-4x-6=2(x^2-2x-3)=2(x+1)(x-3)=0 \Rightarrow x=-1 \text{ or } x=3.\]

Observe that \(x=-1\) is an extraneous root since

\[\text{LHS}=\sqrt{2\cdot(-1)^2+7}=3 \ne 2\cdot (-1)-1=\text{ RHS}.\]

Then since

\[\text{LHS}=\sqrt{2\cdot (3)^2+7}=5=2\cdot (3)-1=\text{ RHS},\]

the solution to the equation \(\sqrt{2x^2+7}+1=2x\) is \(x=3.\) \( _\square \)

Here are the problems for you try.

• Systems of Equations

• Quadratic Equations

• Functional Equations