Solving Radical Equations and Solving Equations Using Radicals

Radical equations are equations containing a radical. To solve radical equations, you will use the opposite operation to "cancel out" the radical, just as you do with other operations.

Therefore, "what you do to one side of an equation, you must do to the other side."

Specifically, when solving radical equations, the following steps are involved:

Isolate the radical (the radical should be the only thing on that side or there should be a radical equal to a radical);
Raise both sides to whatever index number is stated (or the reciprocal of the exponent);
Solve the equation for the given variable;
Check your solutions.

Look at the following examples:

Example One

$LaTeX: \sqrt[]{x}=5 \\ \left(\sqrt[]{x}\right)^2=5^2 \hspace{1cm}\text{You should "square" both sides. When "squaring" the left side in this problem, you should square a square root. }\\ \hspace{4cm}\text{When "squaring" the right side in this problem, you should square a constant.}\\ x=25 \text{ The Final Solution}\\\\ \\ \text {Check your solution} \\ \sqrt[]{25}=5 \\ 5=5 \hspace{3mm} \\ \text{TRUE. Therefore, x=25 is the Final Solution}$

Example Two

You should raise both sides to the 3rd power. When raising the left side to the 3rd power, you are "cubing" a "cube root". When raising the right side to the 3rd power, you are "cubing" a constant.

$LaTeX: \sqrt[3]{x}=3 \\ \left(\sqrt[3]{x}\right)^3=3^3 \hspace{1cm} \text{You should raise both sides to the 3rd power. When raising the left side to the 3rd power, you are "cubing" a "cube root".}\\ \hspace{3cm}\text{When raising the right side to the 3rd power, you are "cubing" a constant.} \\ x=27 \hspace{2cm} \text{The Final Solution}\\ \text{Check your solution}\\ \sqrt[3]{\left(27\right)}=3 \\ 3=\hspace{9mm}3 \\ TRUE \\ \text{Therefore, x=27 is the Final Solution}$

Example Three

$LaTeX: 5-\sqrt[4]{x}=0 \\ -\sqrt[4]{x}=-5 \hspace{2cm}\text{Move the constant term to the right side.}\\ \sqrt[4]{x}=5\hspace{2.6cm}\text{Divide both sides by "-1".}\\ \left(\sqrt[4]{x}\right)^4=\left(5\right)^4 \hspace{1.6cm}\text{Raise both sides to the 4th power.}\\ x=625 \hspace{2.6cm}\text{The Final Solution}\\ \text{Check your solution} \\ 5-\sqrt[4]{\left(625\right)}=0 \\ 5-(5)\hspace{1cm}=0 \\ 0\hspace{2cm}=0 \\ \text{TRUE. Therefore, x=625 is the Final Solution}$

Remember, when you solve radical equations, you should always check your solution(s). Sometimes, when you are checking your answers, there are "extraneous solutions." Extraneous solutions are solutions of the simplified form of the equation that do not satisfy the original equation. Solve the following radical equations and check for extraneous solutions.

Example 1a: Solve equations using nth roots (with one solution)

$LaTeX: 4x^5=128\\ x^5=32 \hspace{1cm}\text{Divide both sides by "4"}\\ x=\sqrt[5]{32} \hspace{.8cm}\text{Take the "5th root" of both sides.}\\ x=2 \hspace{1.3cm}\text{The Final Solution}\\ \text{Test to see if it is an extraneous solution.}\\ 4\left(2\right)^5=128 \\ 4\left(32\right)=128\\ 128=128 \\ \text{TRUE}\\ \text{Therefore, x=2 is the Final Solution.}$

Example 1b: Solve equations using nth roots (with one solution)

$LaTeX: 2\sqrt[]{x+8}=3\sqrt[]{x-2} \\ \left(2\sqrt[]{x+8}\right)^2=\left(3\sqrt[]{x-2}\right)^2 \hspace{1cm} \text{"Square" both sides. When "squaring" the left side in this problem, you square a square root, as well as a square}\\ \hspace{4cm}\text{ a constant term. When "squaring" the right side in this problem, you square a square root, as well as square a constant term.} \\ 4\left(x+8\right)=9\left(x-2\right)\\ 4x+32=9x-18 \hspace{.5cm}\text{Use the Distributive Property to expand the Equation.}\\ -5x=-50 \hspace{.5cm}\text{Move the variable terms to the left side, and move the constant terms to the right side.}\\ x=10 \hspace{.5cm}\text{This is the solution.}\\ \text{Test to see if it is an extraneous solution.} \\ 2\sqrt[]{\left(10\right)+8}=3\sqrt[]{\left(10\right)-2}\\ 2\sqrt[]{18}=3\sqrt[]{8}\\ 2\cdot\sqrt[]{9}\cdot\sqrt[]{2}=3\cdot\sqrt[]{4}\cdot\sqrt[]{2} \\2\cdot3\cdot\sqrt[]{2}=3\cdot2\cdot\sqrt[]{2}\\ 6\sqrt[]{2}=6\sqrt[]{2} \\ \text{TRUE. Therefore, x=10 is the solution.}$

Example 2a: Solve equations using nth roots (with one solution, none extraneous)

$LaTeX: \sqrt[]{5x+3}=\sqrt[]{3x+7}\\ \left(\sqrt[]{5x+3}\right)^2=\left(\sqrt[]{3x+7}\right)^2 \hspace{1cm}\text{"Square" both sides. When "squaring" the left side in this problem, you square a square root. }\\ \hspace{5cm}\text{When "squaring" the right side in this problem, you square a square root.} \\ 5x+3=3x+7 \hspace{1cm}\text{}\\ 2x=4 \hspace{1cm}\text{Move the variable terms to the left side, and move the constant terms to the right side.}\\ x=2 \hspace{1cm}\text{This is the solution.}\\ \text{Test to see if it is an extraneous solution.}\\ \sqrt[]{5\left(2\right)+3}=\sqrt[]{3\left(2\right)+7}\\ \sqrt[]{10+3}=\sqrt[]{6+7}\\ \sqrt[]{13}=\sqrt[]{13}\\ \sqrt[]{13}=\sqrt[]{13} \\ \text{TRUE. Therefore, x=2 is the solution.}$

Example 2b: Solve equations using nth roots (with one solution, but extraneous)

$LaTeX: \sqrt[]{2x-1}+5=2\\ \sqrt[]{2x-1}=\left(-3\right)^2 \hspace{1cm}\text{You should "square" both sides. }\\ \hspace{4cm}\text{When "squaring" the left side in this problem, you square a square root. When "squaring" the right side in this problem,}\\ \hspace{4cm}\text{you square a constant.}\\ 2x-1=9\\ 2x=10 \\ x=5 \\ \text{Test to see if it is an extraneous solution.}\\ \sqrt[]{2\left(5\right)-1}+5=2\\ \sqrt[]{9}+5=2\\ 3+5=2\\ 8\ne2 \\ \text{FALSE. Because x=5 creates a false statement, x≠5, x=5 is an extraneous solution. Therefore, there is No Solution.}$

Example 2c: Solve equations using nth roots (with one solution, but extraneous)

$LaTeX: \sqrt[4]{8x+12}-9=-11\\ \sqrt[4]{8x+12}=-2 \hspace{1cm}\text{Add "9" to both sides.} \left(\sqrt[4]{8x+12}\right)^4=\left(-2\right)^4\\ 8x+12=16 \\ 8x=4 \hspace{2cm}\text{Subtract "4" from both sides.} \\ x=\frac{1}{2} \hspace{2cm}\text{Divide both sides by "8"}\\ x=\frac{1}{2} \hspace{2cm}\text{This is the Solution.} \\ \text{Test to see if it is an extraneous solution.}\\ \sqrt[4]{8\left(\frac{1}{2}\right)+12}-9=-11\\ \sqrt[4]{4+12}-9=-11\\ \sqrt[4]{16}-9=-11 \\ 2-9=-11 \\ -7=-11 \\ -7\ne -11 \\ \text{FALSE. Therefore, there is No Solution.}$

Example 3a: Solve equations using nth roots (with two solutions)

$LaTeX: \left(x-3\right)^4=21\\ \sqrt[4]{\left(\left(x-3\right)\right)}=\sqrt[4]{21} \hspace{1cm}\text{Take the 4th Root of both sides}\\ x=3\pm\sqrt[4]{21}\hspace{1cm}\text{When taking the 4th Root of both sides, the }\pm\text{of the radical of a number must take place.}\\ x=3\pm\sqrt[4]{21}\hspace{1cm} \text{The Solution}\\ \text{or}\\ x=3+\sqrt[4]{21},\:3-\sqrt[4]{21} \hspace{1cm}\text{The Solution may also be written as two separate solutions.}\\ x\approx 5.14 \text{or}x\approx 0.86 \hspace{1cm}\text{The Solution may also be written as two separate decimal approximations.}\\ \text{Test for extraneous solutions:}\\ \left(\left(5.14-3\right)^4\right)=21\:or\:\left(\left(0.86-3\right)^4\right)=21\\ (2.14)^4=21\:or\:(-2.14)^4=21\\ 20.973\approx 21\:or\hspace{.6cm}\:20.973\approx 21\\ TRUE \hspace{2cm}TRUE\\ x=3+\sqrt[4]{21}\hspace{1.6cm}3-\sqrt[4]{21}\\ or\\ x\approx 5.14\hspace{2cm}0.86$

Example 3b: Solve equations using nth roots (with two solutions, but one is extraneous)

$LaTeX: x=1+\sqrt[]{5x+9}\\ x=1+\sqrt[]{5x+9}\hspace{1cm}\text{Subtract "1" from both sides.}\\ \left(x-1\right)^2=\left(\sqrt[]{5x+9}\right)^2\hspace{1cm}\text{You should "square" both sides.}\\ \hspace{3cm}\text{When "squaring" the left side in this problem, you should square a binomial.}\\ \hspace{3cm}\text{When "squaring" the right side in this problem, you should square a square root.}\\ x^2-2x+1=5x+9\hspace{1cm}\text{Expand the binomial on the left side.}\\ x^2-7x-8=0\hspace{1cm}\text{Move the two terms to the left by subtraction.}\\ (x-8)(x+1)=0\hspace{1cm}\text{Factor into two binomials.}\\ (x-8)=0\:or\:(x+1)=0\hspace{1cm}\text{Set each binomial equal to "0".}\\ x=9\:or\:x=-1\\ \text{Text for extraneous solutions:}\\ \left(8\right)=1+\sqrt[]{5\left(8\right)+9}\:or\:\left(-1\right)=1+\sqrt[]{5\left(-1\right)+9}\\8=1+\sqrt[]{49}\:or\:\left(-1\right)=1+\sqrt[]{4}\\ 8=1+7\hspace{.7cm}or\:(-1)=1+2\\ 8=8\hspace{1.2cm}or\hspace{.8cm}(-1)=3\\ TRUE\hspace{2cm}FALSE\hspace{1cm}\text{Because x=-1 creates a false statement,}\\ x=8\hspace{2cm}x≠-1\hspace{1cm}\text{x=-1 is extraneous solution, and therefore, x=-1 is NOT a solution.}$

Example 3c: Solve equations using nth roots (with two solutions, but one is extraneous)

$LaTeX: x-3=\sqrt[]{30-2x}\\ \left(x-3\right)^2=\left(\sqrt[]{30-2x}\right)\hspace{1cm}\text{"Square" both sides.}\\ \hspace{3cm}\text{When "squaring" the left side in this problem, you square a binomial.}\\ \hspace{3cm}\text{When "squaring" the right side in this problem, you square a square root.}\\ x^2-6x+9=30-2x\hspace{1cm}\text{Expand the binomial on the left side.}\\ x^2-4x-21=0\hspace{1cm}\text{Move the two terms to the left side.}\\ \left(x-7\right)\left(x+3\right)=0\hspace{1cm}\text{Factor into two binomials. }\\ \left(x-7\right)=0\:or\:\left(x+3\right)\:=0\hspace{1cm}\text{Set binomial equal to "0".}\\ x=7\:or\:x=-3\hspace{1cm}\text{These are the two solutions.}\\ \text{Test for extraneous solutions:}\\ 7-3=\sqrt[]{30-2\left(7\right)}\:\:\:or-3-3=\sqrt[]{20-2\left(-3\right)}\\ 4=\sqrt[]{30-14}\hspace{2cm}or\:-6=\sqrt[]{30+6}\\ 4=\sqrt[]{16}\hspace{2.5cm}or\:-6=\sqrt[]{36}\\ 4=4\hspace{2cm}or\:-6=6\\ 4=4\hspace{2cm}or\:-6=6\hspace{1cm}\\ TRUE\hspace{2cm}FALSE\hspace{1cm}\text{Because x=-3 creates a false statement,}\\ x=7\hspace{2cm}x≠-3\hspace{1cm}\text{x-3 is an extraneous solution, and }\\ \text{therefore, x=-3 is NOT a solution.}$

Example 4a: Use nth roots in problem solving

A study determined that the weight w (in grams) of coral code near Palawan Island, Philippines, can be approximated using the model $LaTeX: w=0.0167\left(l\right)^3$ where l is the coral cod's length (in centimeters). Estimate the length of a coral code that weights 200 grams.

Solution

$LaTeX: w=0.0167\left(l\right)^3\\ 200=0.0167\left(l\right)^3\\ 200=\frac{0.0167\left(l\right)^3}{0.0167}\\ 11,976=l^3\\ \sqrt[3]{11,976}\approx l\\ l\approx 22.9\\ \text{Therefore, a coral cod that weighs 200 grams is approximately 23 cm long}\\ \text{Test to see if it is an extraneous solution.}\\ 200=0.0167(22.9)^3\\ 200=0.0167\cdot (12,008,989) \\ 200\approx 200.5501\\ TRUE\\ l\approx 22.9,\text{which rounds to 23 cm}$

Example 4b: Use nth roots in problem solving

At a certain electronics company, the daily output Q is related to the number of people A on the assembly line by $LaTeX: Q=400+\sqrt[]{A+25}$ . Determine how many people are needed on the assembly line if the daily output is to be 525.

$LaTeX: 525=500+\sqrt[]{A+25}\\ 125=\sqrt[]{A+25}\hspace{1cm}\text{Move the constant term to the right side.}\\ \left(125\right)^2=\left(\sqrt[]{A+25}^{ }\right)\hspace{1cm}\text{"Square" both sides. When "squaring" the left side in this problem, you square a constant term.}\\ \hspace{4cm}\text{When "squaring" the right side in this problem, you square a square root.}\\ 15,600=A\\ A=15,600\hspace{1cm}\text{This is the solution.}\\ \text{Test to see if it is an extraneous solution.}\\ 525=400+\sqrt[]{\left(15,600\right)+25}\\ 525=400\sqrt[]{15,625}\\ 525=400+125\\ 525=525\\ TRUE$

Solving Radical Equations Presentation

Solving Radical Equations Practice

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