Solve equations with radicals and check for extraneous solution

Here we look at equations that have roots in the problem. As you might expect, to clear a root we can raise both sides to an exponent. So to clear a square root we can rise both sides to the second power. To clear a cubed root we can raise both sides to a third power. There is one catch to solving a problem with roots in it, sometimes we end up with solutions that do not actually work in the equation. This will only happen if the index on the root is even, and it will not happen all the time. So for these problems it will be required that we check our answer in the original problem. If a value does not work it is called an extraneous solution and not included in the final solution.

When solving a radical problem with an even index: check answers!

Example 1:

Also Read:

Solve \sqrt{7x+2}=4

Answer:

Even index! We will have to check answers

\sqrt{7x+2}=4

Square both sides, simplify exponents

(\sqrt{7x+2})^2=4^2

7x +2 = 16

Solve

7x + 2 – 2 = 16 – 2

7x = 14

\frac{7x}{7}=\frac{14}{7}

x = 2.

Need to check answer in original problem

\sqrt{7(2)+2}=4

\sqrt{14+2}=4

\sqrt{16}=4

4 = 4

True! It works!, so the solution is x = 2.

Example 2:

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

Answer:

Odd index, we don’t need to check answer

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

Cube both sides, simplify exponents

(\sqrt[3]{x-1})^3=(-4)^3

x – 1 = – 64

x – 1 + 1 = – 64 + 1

x = – 63 ( our solution).

Example 3:

Solve \sqrt[4]{3x+6}=-3

Answer:

Even index! We will have to check answers

\sqrt[4]{3x+6}=-3

Rise both sides to fourth power

(\sqrt[4]{3x+6})^4=(-3)^4

3x + 6 = 81

Solve this equation

3x + 6 – 6 = 81 – 6

3x = 75

x = 25

Need to check answer in original problem

\sqrt[4]{3x+6}=-3

\sqrt[4]{3(25)+6}=-3

\sqrt[4]{75+6}=-3

\sqrt[4]{81}=-3

3 = -3 False, extraneous solution

Our solution is no solution.

If the radical is not alone on one side of the equation we will have to solve for the radical before we raise it to an exponent

Example 4:

Solve x+\sqrt{4x+1}=5

Answer:

Even index! We will have to check solutions

x+\sqrt{4x+1}=5

Isolate radical by subtracting x from both sides

\sqrt{4x+1}=5-x

Square both sides

(\sqrt{4x+1})^2=(5-x)^2

Evaluate exponents, recal (a-b)^2=a^2-2ab+b^2

4x+1=25-10x+x^2

Re -order terms

4x+1=x^2-10x+25

Make equation equal zero

0=x^2-14x+24

Factor

0 =(x – 12)(x – 2)

Set each factor equal to zero

x – 12 = 0 or x – 2 = 0

Solve each equation

x = 12 or x = 2

Need to check answers in original problem

Check x = 12 first

12+\sqrt{4(12)+1}=5

12+\sqrt{48+1}=5

12+\sqrt{49}=5

12+7=5

19 = 5 False, extraneous root

Check x =2

x+\sqrt{4x+1}=5

2+\sqrt{4(2)+1}=5

2+\sqrt{8+1}=5

2+\sqrt{9}=5

2 + 3 = 5

5 = 5 True! It works

Our solution is x = 2.

The above example illustrates that as we solve we could end up with an x2 term or a quadratic. In this case we remember to set the equation to zero and solve by factoring. We will have to check both solutions if the index in the problem was even. Sometimes both values work, sometimes only one, and sometimes neither works.

World View Note: The Babylonians were the first known culture to solve quadratics in radicals – as early as 2000 BC!

If there is more than one square root in a problem we will clear the roots one at a time. This means we must first isolate one of them before we square both sides.

Example 5:

Solve \sqrt{3x-8}-\sqrt{x}=0

Answer:

\sqrt{3x-8}-\sqrt{x}=0

Even index! We will have to check answers

Isolate first root by adding \sqrt{x} to both sides

\sqrt{3x-8}=\sqrt{x}

Square both sides

(\sqrt{3x-8})^2=(\sqrt{x})^2

3x – 8 = x

3x – x = 8

2x = 8

x = 4

Need to check answer in original

\sqrt{3x-8}-\sqrt{x}=0

\sqrt{3(4)-8}-\sqrt{4}=0

\sqrt{12-8}-\sqrt{4}=0

\sqrt{4}-\sqrt{4}=0

2 – 2 = 0 True! It works

Our solution is x = 4.

Also Read : Solve equations that are quadratic in form by substitution to create a quadratic equation

When there is more than one square root in the problem, after isolating one root and squaring both sides we may still have a root remaining in the problem. In this case we will again isolate the term with the second root and square both sides. When isolating, we will isolate the term with the square root. This means the square root can be multiplied by a number after isolating.

Example 6:

Solve \sqrt{2x+1}-\sqrt{x}=1

Answer:

\sqrt{2x+1}-\sqrt{x}=1

Even index! We will have to check answers

Isolate first root by adding \sqrt{x} to both sides

\sqrt{2x+1}=\sqrt{x}+1

Square both sides

(\sqrt{2x+1})^2=(\sqrt{x}+1)^2

2x+1=x+2\sqrt{x}+1

x=2\sqrt{x}

Square both sides

x^2=(2\sqrt{x})^2

x2 = 4x

x2 – 4x = 0

x ( x – 4) = 0

x = 0 or x – 4 = 0

x = 0 or x = 4

Need to check answers in original

x = 0

\sqrt{2(0)+1}-\sqrt{0}=1

\sqrt{1}-\sqrt{0}=1

1 – 0 = 1

1 = 1 True! It works

Check x = 4

\sqrt{2(4)+1}-\sqrt{4}=1

\sqrt{8+1}-\sqrt{4}=1

\sqrt{9}-\sqrt{4}=1

3 – 2 = 1

1 = 1 True! It works

Our solution is x = 0 or x = 4.

 

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