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5. Rational Exponents & Radical Functions
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United States of AmericaFL
Grade 912

5.05 Solving radical equations

Lesson
Practice
Lesson

Concept summary

Radical equation

An equation containing at least one radical expression

The process of solving radical equations is essentially the same as solving a linear equation, that is, by using inverse operations. The inverse of a square root operation is squaring, and the inverse of an nth root, is to take the nth power.

It is important to remember that the parent function f\left(x\right)=\sqrt{x} has a domain of x\geq0, and a range of y\geq0. When we are solving equations involving square roots, and other even roots, we need to remember that on every step of our solution, any radicands must be non-negative and any radicals must also be non-negative. It is important to note, however, that the solution to the equation itself can be negative.

When we square both sides we may get a solution that results in a negative radicand, or a solution which will result in a false statement if substituted into the original equation. These are known as extraneous solutions.

When solving questions with real life applications, we also need to ensure we have viable solutions, which make sense within the context of the question. A non-viable solution does not make sense within the context of the question, such as a negative value when we are solving for the length of a physical object.

Worked examples

Example 1

Solve the following equations:

a

\sqrt{y}+5=9

Approach

To solve for y, we first want to isolate \sqrt{y}, we can then square both sides of the equation.

Solution

\displaystyle \sqrt{y}+5 \displaystyle =\displaystyle 9State the equation
\displaystyle \sqrt{y} \displaystyle =\displaystyle 4Subtract 5 from both sides
\displaystyle \left(\sqrt{y}\right)^2\displaystyle =\displaystyle 4^2Square both sides
\displaystyle y\displaystyle =\displaystyle 16Evaluate the squares

Reflection

We can see that if we substitute y=16 into the original equation we obtain \sqrt{16}+5=9 which is a true statement, and as \sqrt{16} is positive we have a valid solution.

b

\sqrt[3]{4x-9}=-1

Approach

To solve for x, we first want to isolate 4x-9 by raising each side of the equation to a power of 3. We are then left with a linear equation we can solve using more inverse operations

Solution

\displaystyle \sqrt[3]{4x-9} \displaystyle =\displaystyle -1State the equation
\displaystyle \left(\sqrt[3]{4x-9}\right)^3 \displaystyle =\displaystyle \left(-1\right)^3Raise both sides to a power of 3
\displaystyle 4x-9\displaystyle =\displaystyle -1Evaluate the cubes
\displaystyle 4x\displaystyle =\displaystyle 8Add 9 to both sides
\displaystyle x\displaystyle =\displaystyle 2Divide both sides by 4

Reflection

We can see that if we substitute x=2 into the original equation we obtain \sqrt[3]{-1}=-1 which is a true statement. When dealing with odd valued indexes such as \sqrt[3]{x}, we do not need to worry about the radicand being negative.

Raising both sides of an equation to an odd power is reversible. If a=b then a^3=b^3 and if a^3=b^3 then a=b.

Example 2

Solve each equation for x. Identify any extraneous solutions.

a

3x=1+2\sqrt{x}

Approach

To solve for x, we first want to isolate 2\sqrt{x}. We can then square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

Solution

\displaystyle 3x\displaystyle =\displaystyle 1+2\sqrt{x}State the equation
\displaystyle 3x-1\displaystyle =\displaystyle 2\sqrt{x}Subtract 1 from both sides
\displaystyle \left(3x-1\right)^2\displaystyle =\displaystyle \left(2\sqrt{x}\right)^2Square both sides
\displaystyle 9x^2-6x+1\displaystyle =\displaystyle 4xEvaluate the squares
\displaystyle 9x^2-10x+1\displaystyle =\displaystyle 0Subtract 4x from both sides
\displaystyle \left(9x-1\right)\left(x-1\right)\displaystyle =\displaystyle 0Factor the quadratic

Using the zero product rule we get two solutions x=\dfrac{1}{9}, x=1. We now want to test both solutions:

\displaystyle 3\left(1\right)\displaystyle =\displaystyle 1+2\sqrt{1}Substitute x=1
\displaystyle 3\displaystyle =\displaystyle 3Evaluate both sides

We can see that x=1 satisfies the original equation and is a valid solution.

Now, let's test x=\dfrac{1}{9}

\displaystyle 3\left(\dfrac{1}{9}\right)\displaystyle =\displaystyle 1+2\sqrt{\frac{1}{9}}Substitute x=\dfrac{1}{9}
\displaystyle \frac{1}{3}\displaystyle =\displaystyle \frac{5}{3}Evaluate both sides

We can see that x=\dfrac{1}{9} leads to a false statement and is therefore an extraneous solution.

b

\sqrt{x+17}=x+5

Approach

To solve for x, we first want to square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

Solution

\displaystyle \sqrt{x+17}\displaystyle =\displaystyle x+5State the equation
\displaystyle \left(\sqrt{x+17}\right)^2\displaystyle =\displaystyle \left(x+5\right)^2Square both sides
\displaystyle x+17\displaystyle =\displaystyle x^2+10x+25Evaluate the squares
\displaystyle 0\displaystyle =\displaystyle x^2+9x+8Subtract x+17 from both sides
\displaystyle 0\displaystyle =\displaystyle \left(x+1\right)\left(x+8\right)Factor the quadratic

Using the zero product rule we get two solutions x=-8, x=-1. We now want to test both solutions:

\displaystyle \sqrt{-8+17}\displaystyle =\displaystyle -8+5Substitute x=-8
\displaystyle 3\displaystyle =\displaystyle -3Evaluate both sides

We can see that x=-8 leads to a false statement and is therefore an extraneous solution.

Now, let's test x=-1.

\displaystyle \sqrt{-1+17}\displaystyle =\displaystyle -1+5Substitute x=-1
\displaystyle 4\displaystyle =\displaystyle 4Evaluate both sides

We can see that x=-1 satisfies the original equation and is a valid solution.

Reflection

It is important to note that the fact that our solutions were negative, x=-8, x=-1, does not necessarily make them extraneous. We can see after substitution that only x=- 8 is an extraneous solution, and x=-1 is valid.

Example 3

The radius r of a cone whose height is equal to twice its radius is given by r = \sqrt[3]{\dfrac{3V}{2 \pi}}. Carlos is studying an underwater volcano that is roughly this shape. Solve for the volume, V, of the volcano if it has a radius of 1.3\text{ km}. Give your answer in km^3, rounded to one decimal place.

Solution

\displaystyle r\displaystyle =\displaystyle \sqrt[3]{\dfrac{3V}{2 \pi}}State the equation
\displaystyle 1.3\displaystyle =\displaystyle \sqrt[3]{\dfrac{3V}{2 \pi}}Substitute r=1.3
\displaystyle 1.3^3\displaystyle =\displaystyle \left(\sqrt[3]{\dfrac{3V}{2 \pi}}\right)^3Raise both sides to a power of 3
\displaystyle 2.197\displaystyle =\displaystyle \dfrac{3V}{2 \pi}Evaluate the cubes
\displaystyle 2.197\left(2 \pi\right)\displaystyle =\displaystyle 3VMultiply both sides by 2\pi
\displaystyle \dfrac{2.197\left(2\pi\right)}{3}\displaystyle =\displaystyle VDivide both sides by 3
\displaystyle 4.6\displaystyle \approx\displaystyle VEvaluate, rounding to one decimal place

The volcano will have a volume of 4.6 km^3.

Reflection

We can also solve this by rearranging the given equation before substituting in the known value.

\displaystyle V\displaystyle =\displaystyle \dfrac{2\pi r^3}{3}
\displaystyle V\displaystyle =\displaystyle \dfrac{2\pi \left(1.3^3\right)}{3}
\displaystyle V\displaystyle \approx\displaystyle 4.6

Outcomes

MA.912.AR.7.1

Solve one-variable radical equations. Interpret solutions as viable in terms of context and identify any extraneous solutions.

MA.912.AR.7.3

Solve and graph mathematical and real-world problems that are modeled with square root or cube root functions. Interpret key features and determine constraints in terms of the context.

MA.912.NSO.1.3

Generate equivalent algebraic expressions involving radicals or rational exponents using the properties of exponents.

MA.912.NSO.1.5

Add, subtract, multiply and divide algebraic expressions involving radicals.

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