Lesson Objectives
- Demonstrate an understanding of radical expressions
- Learn how to solve an equation with radicals
- Learn how to identify extraneous solutions
How to Solve Radical Equations
When we solve equations with radicals, we need a new rule. When both sides of an equation are raised to the same power, all solutions of the original equation are solutions to the new equation. We must be careful here, each solution of the new equation is not necessarily a solution to the original. When working with radical equations, we must always check for extraneous solutions or solutions that do not work in the original equation. To see this in action, let's think about a very simple example.
x = 7
Let's suppose we square both sides of the above equation:
x2 = 49
This new equation has two solutions, x can be 7 or (-7).
(7)2 = 49
(-7)2 = 49
-7 does not work as a solution in our original equation.
-7 = 7 (false)
This is an extraneous solution and occurred from the squaring operation.
Example 1: Solve each equation. $$\require{color}\sqrt{2x - 7}=1$$ Step 1) Isolate one of the radicals.
In this case, our radical is already isolated on the left side of the equation. $$\sqrt{2x - 7}=1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation: $$\left(\sqrt{2x - 7}\right)^2=1^2$$ $$2x - 7=1$$ Step 3) Solve the equation. $$2x - 7=1$$ $$2x=8$$ $$x=4$$ Step 4) Check all solutions in the original equation. $$\sqrt{2x - 7}=1$$ $$\sqrt{2(4) - 7}=1$$ $$\sqrt{8 - 7}=1$$ $$\sqrt{1}=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Example 2: Solve each equation. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ Step 1) Isolate one of the radicals.
We will isolate the leftmost radical: $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation. $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ $$\left(\sqrt{4x + 1}\right)^2=\left(\sqrt{2x + 4}+ 1\right)^2$$ $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ In this case, we still have a radical involved. We will repeat our first two steps.
Step 1) Isolate one of the radicals.
Let's isolate our radical on the right side of the equation: $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ $$2x - 4=2\sqrt{2x + 4}$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation. $$\left(2x - 4\right)^2=\left(2\sqrt{2x + 4}\right)^2$$ $$4x^2 - 16x + 16=4(2x + 4)$$ $$4x^2 - 16x + 16=8x + 16$$ $$4x^2 - 24x=0$$ Step 3) Solve the equation. $$4x^2 - 24x=0$$ $$4x(x - 6)=0$$ $$4x=0$$ $$x=0$$ $$x - 6=0$$ $$x=6$$ Step 4) Check all solutions in the original equation.
We have two proposed solutions, x = 6 and x = 0. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(0) + 1}- \sqrt{2(0) + 4}=1$$ $$\sqrt{1}- \sqrt{4}=1$$ $$1 - 2=1$$ $$-1=1 \hspace{.2em}\color{red}{✖}$$ $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(6) + 1}- \sqrt{2(6) + 4}=1$$ $$\sqrt{24 + 1}- \sqrt{12 + 4}=1$$ $$\sqrt{25}- \sqrt{16}=1$$ $$5 - 4=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Our solution of x = 0 does not work in the original equation, therefore, this is an extraneous solution.
Our only valid solution is x = 6.
x = 7
Let's suppose we square both sides of the above equation:
x2 = 49
This new equation has two solutions, x can be 7 or (-7).
(7)2 = 49
(-7)2 = 49
-7 does not work as a solution in our original equation.
-7 = 7 (false)
This is an extraneous solution and occurred from the squaring operation.
Solving Equations with Radicals
- Isolate one of the radicals
- Raise both sides of the equation to a power equal to the index of the radical
- Repeat the previous two steps if necessary
- Solve the equation
- Check all solutions in the original equation
Example 1: Solve each equation. $$\require{color}\sqrt{2x - 7}=1$$ Step 1) Isolate one of the radicals.
In this case, our radical is already isolated on the left side of the equation. $$\sqrt{2x - 7}=1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation: $$\left(\sqrt{2x - 7}\right)^2=1^2$$ $$2x - 7=1$$ Step 3) Solve the equation. $$2x - 7=1$$ $$2x=8$$ $$x=4$$ Step 4) Check all solutions in the original equation. $$\sqrt{2x - 7}=1$$ $$\sqrt{2(4) - 7}=1$$ $$\sqrt{8 - 7}=1$$ $$\sqrt{1}=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Example 2: Solve each equation. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ Step 1) Isolate one of the radicals.
We will isolate the leftmost radical: $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation. $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ $$\left(\sqrt{4x + 1}\right)^2=\left(\sqrt{2x + 4}+ 1\right)^2$$ $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ In this case, we still have a radical involved. We will repeat our first two steps.
Step 1) Isolate one of the radicals.
Let's isolate our radical on the right side of the equation: $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ $$2x - 4=2\sqrt{2x + 4}$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation. $$\left(2x - 4\right)^2=\left(2\sqrt{2x + 4}\right)^2$$ $$4x^2 - 16x + 16=4(2x + 4)$$ $$4x^2 - 16x + 16=8x + 16$$ $$4x^2 - 24x=0$$ Step 3) Solve the equation. $$4x^2 - 24x=0$$ $$4x(x - 6)=0$$ $$4x=0$$ $$x=0$$ $$x - 6=0$$ $$x=6$$ Step 4) Check all solutions in the original equation.
We have two proposed solutions, x = 6 and x = 0. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(0) + 1}- \sqrt{2(0) + 4}=1$$ $$\sqrt{1}- \sqrt{4}=1$$ $$1 - 2=1$$ $$-1=1 \hspace{.2em}\color{red}{✖}$$ $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(6) + 1}- \sqrt{2(6) + 4}=1$$ $$\sqrt{24 + 1}- \sqrt{12 + 4}=1$$ $$\sqrt{25}- \sqrt{16}=1$$ $$5 - 4=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Our solution of x = 0 does not work in the original equation, therefore, this is an extraneous solution.
Our only valid solution is x = 6.
Skills Check:
Example #1
Solve each equation. $$1 + \sqrt{4 - x}=\sqrt{3x - 5}$$
Please choose the best answer.
A
$$x=3, 5$$
B
$$x=-8, 3$$
C
$$x=-2$$
D
$$x=3$$
E
No Solution
Example #2
Solve each equation. $$\sqrt{x + 3}=x - 3$$
Please choose the best answer.
A
$$x=-1$$
B
$$x=6$$
C
$$x=-\frac{1}{2}, 3$$
D
$$x=-4, 7$$
E
$$x=-5, \frac{13}{2}$$
Example #3
Solve each equation. $$\sqrt{6 - x}+ 1=\sqrt{9 - x}$$
Please choose the best answer.
A
$$x=2$$
B
$$x=-\frac{1}{2}, 9$$
C
$$x=-\frac{3}{4}, 7$$
D
$$x=5$$
E
$$x=-5, 2$$
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