Lesson Solving equations containing radicals

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How to solve equations containing radicals

An equation is called a radical equation if it contains the variable under a square root, cube root etc.

There are two major methods for solving equations containing radicals:
1) To isolate radical on one side of the equation and then to eliminate it by raising both sides to a power equal to the radical index, and
2) To introduce a new variable.

Examples below show how to use these methods.

Example 1

Solve the equation root%283%2C2x-3%29+=+3

.

Solution
Raise each side to a 3-rd power (the index of the radical is 3). You get the equation
2x-3+=+27

.
It has a solution x=15

.
You should check this solution. To check it, simply substitute the value x=15

into the original equation. You get
root%283%2C2%2A15-3%29+=+root%283%2C27%29+=+3

.
So, the found solution x+=15

is correct.

Exercise 1
Solve yourself the equation root%286%2Cx-3%29+=+2

Example 2

Solve the equation sqrt%28x-1%29+%2B+sqrt%282x%2B6%29+=+6

.

Solution
Let us isolate the radical expression sqrt%282x%2B6%29

on the left side:
sqrt%282x%2B6%29+=+6+-+sqrt%28x-1%29

.
Now raise both sides to a degree 2 (the index of the radical is 2). You get
%28sqrt%282x%2B6%29%29%5E2+=+%286+-+sqrt%28x-1%29%29%5E2

2x%2B6+=+36+-+2%2A6%2Asqrt%28x-1%29+%2B+x-1

,
and after simplifications
12%2Asqrt%28x-1%29+=+29-x

.
Since the equation still contains radical (now on the left side only), raise both sides to a degree 2 one more time:
144%2A%28x-1%29+=+%2829-x%29%5E2

.
Open brackets, collect all common terms on one side. You get
144x-144+=+841+-+58x+%2Bx%5E2

.
The last equation is the quadratic equation.
Apply quadratic formula (see the lesson Introduction into Quadratic Equations)
x+=+%28202%2B-sqrt%28202%5E2-4%2A985%29%29%2F2+=+%28202%2B-sqrt%2840804-3940%29%29%2F2+=+%28202%2B-sqrt%2836864%29%29%2F2

x+=+%28202%2B-sqrt%28202%5E2-4%2A985%29%29%2F2+=+%28202%2B-sqrt%2840804-3940%29%29%2F2+=+%28202%2B-sqrt%2836864%29%29%2F2

.
Two roots of this equation are x%5B1%5D+=+5

and

.
To check it, simply substitute these values into the original equation.
For x=5

you have

sqrt%285-1%29%2Bsqrt%282%2A5%2B6%29+=+sqrt%284%29%2Bsqrt%2816%29+=+2%2B4=6

.
This means x=5

is the solution of the original equation.
For x=197

you have

sqrt%28197-1%29%2Bsqrt%282%2A197%2B6%29+=+sqrt%28196%29%2Bsqrt%28400%29+=+14%2B20+=+34

.
It is not equal to 6, so x=197

is not a solution. This is the extraneous solution.

Answer: x=5

.

Exercise 2
Solve yourself the equation sqrt%282x%2B3%29+-+sqrt%28x%2B2%29+=+2

Example 3

Solve the equation root%282%2F3%2Cx-2%29+-+root%281%2F3%2Cx-2%29+=+2

.

Solution
Let us apply the method that introduces a new variable.
Put y=root%281%2F3%2Cx-2%29

. Then

, and you can rewrite the given equation in the form
y%5E2+-+y+-+2+=+0

.
This is the quadratic equation. It has roots
y=2

and

.
Now the problem is to solve the equations root%281%2F3%2Cx-2%29+=+2

and

.
Raising both sides of the equation root%281%2F3%2Cx-2%29+=+2

to degree 3 (the index of the radical is 3) you get x-2=2%5E3=8

. Hence,

.
The second equation root%281%2F3%2Cx-2%29+=+-1

has no solution in real numbers, because the domain for the fractional power is the set of positive real numbers, but the fractional power of the positive real number is the positive real number again.

Answer: x=10

.

Exercise 3
Solve yourself the equation root%282%2F5%2Cx-5%29-2%2Aroot%281%2F5%2Cx-5%29+=+3

introducing new variable.

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