Radical terms can appear in a number of expressions and equations in Math, which makes knowing how to solve equations with radicals quite important.
At a more basic level, equations with radicals that need to be solved will mostly feature just the square root of a term or expression.
These types of radical equations are the kind that will be featured on this page, as we give an introduction to how to solve equations with radicals involved.
Solve Equations with Radicals Intro:
A standard radical equation we can consider is, \boldsymbol{x = \sqrt{4x + 12}}.
We can attempt to solve for x.
The first step is to look at getting rid of the radical in the equation, squaring both sides will do this here.
Then we can rewrite a bit to obtain x values.
x = \sqrt{4x + 12} => x^2 = 4x + 12
x^2 \space {\text{--}} \space 4x \space {\text{--}} \space 12 = 0
(x + 2)(x \space {\text{--}} \space 6) = 0
x = {\text{-}}2 \space\space , \space\space x = 6
So we have two values that we now need to CHECK by inputting them into the original radical equation to see if they hold.
\underline{x = {\text{-}}2}
{\text{-}}2 = \sqrt{4({\text{-}}2) + 12} => {\text{-}}2 = \sqrt{4} => {\text{-}}2 = 2
This is incorrect, so x = {\text{-}}2 is NOT a solution to the radical equation.
\underline{x = 6}
6 = \sqrt{4(6) + 12} => 6 = \sqrt{36} => 6 = 6
This is correct, so x = 6 is a solution to the radical equation.
This is how we can approach finding the correct solution or solutions to basic equations with radicals.
Equations with Radicals Examples
(1.1)
Solve for a. a \space {\text{--}} \space \sqrt{a + 15} = 5
Solution
We can rewrite the equation to get the radical on its own, then square both sides.
a \space {\text{--}} \space 5 \space = \space \sqrt{a + 15} => (a \space {\text{--}} \space 5)^2 = a + 15
[ (a \space {\text{--}} \space 5)^2 \space = \space (a \space {\text{--}} \space 5)(a \space {\text{--}} \space 5) \space = \space a^2 \space {\text{--}} \space 10a + 25 ]
a^2 \space {\text{--}} \space 10a + 25 \space = \space a + 15
a^2 \space {\text{--}} \space 11a + 10 \space = \space 0
(a \space {\text{--}} \space 10)(a \space {\text{--}} \space 1) = 0
a = 10 \space\space , \space\space a = 1
Now to CHECK.
\underline{a = 1}
1 \space {\text{--}} \space \sqrt{1 + 15} = 5 => 1 \space {\text{--}} \space 4 = 5 => {\text{-}}3 = 5 Incorrect
\underline{a = 10}
10 \space {\text{--}} \space \sqrt{10 + 15} = 5 => 10 \space {\text{--}} \space 5 = 5 => 5 = 5 Correct
Solution is a = 5
(1.2)
Solve for b. \sqrt{2b + 3} \space = \space b + 2
Solution
Again we start off by squaring both sides to eliminate the radical.
2b + 3 \space = \space (b + 2)^2
[ (b + 2)^2 \space = \space (b + 2)(b + 2) \space = \space b^2 + 4b + 4 ]
2b + 3 \space = \space b^2 + 4b + 4
b^2 + 2b + 1 \space = \space 0
(b + 1)(b + 1) \space = \space 0
b = {\text{-}}1
Only one value found to CHECK here.
\underline{b = {\text{-}}1}
\sqrt{2({\text{-}}1) + 3} \space = \space {\text{-}}1 + 2 => \sqrt{1} = 1 => 1 = 1 Correct
Solution is b = –1