If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Radical and rational equations | Lesson

## What are radical and rational equations, and how frequently do they appear on the test?

Radical equations are equations in which variables appear under radical symbols (square root of, empty space, end square root).
• square root of, 2, x, minus, 1, end square root, equals, x is a radical equation.
Rational equations are equations in which variables can be found in the denominators of rational expressions.
• start fraction, 1, divided by, x, plus, 1, end fraction, equals, start fraction, 2, divided by, x, end fraction is a rational equation.
Both radical and rational equations can have extraneous solutions, algebraic solutions that emerge as we solve the equations that do not satisfy the original equations. In other words, extraneous solutions seem like they're solutions, but they aren't. On the SAT, we sometimes need to identify and exclude extraneous solutions from the solution set by plugging solutions back into the original equation.
On your official SAT, you'll likely see 0 to 4 questions that test your ability to solve radical and rational equations and identify any extraneous solutions.
You can learn anything. Let's do this!

## How do I solve radical equations?

### Intro to square-root equations & extraneous solutions

Intro to square-root equations & extraneous solutionsSee video transcript

### What do I need to know to solve radical equations?

The process of solving radical equations almost always involves rearranging the radical equations into
, then solving the quadratic equations. As such, knowledge of how to manipulate polynomials algebraically and solve a variety of quadratic equations is essential to successfully solving radical equations.
1. Isolate the radical expression to one side of the equation.
2. Square both sides the equation.
3. Rearrange and solve the resulting equation.

Example: If square root of, 2, x, minus, 1, end square root, equals, x, what is the value of x ?

When it comes to extraneous solutions, the concept that confuses the most students is that of the principal square root. The square root operation gives us only the principal square root, or positive positive square root. For example, square root of, 4, end square root, equals, 2, not both minus, 2 and 2 even though left parenthesis, minus, 2, right parenthesis, squared, equals, 2, squared, equals, 4. If a solution leads to equating the square root of a number to a negative number, then that solution is extraneous.
To check for extraneous solutions to a radical equation:
1. Solve the radical equation as outlined above.
2. Substitute the solutions into the original equation. A solution is extraneous if it does not satisfy the original equation.

Example: What is the solution to the equation square root of, 3, x, plus, 4, end square root, equals, x ?

### Try it!

Try: identify the steps to solving a radical equation
square root of, 2, x, minus, 9, end square root, equals, x, minus, 6
To solve the equation above, we first
both sides of the equation, then rewrite the result as a
equation. Solving this equation gives us 2 solutions, and we
check for extraneous solutions.

Try: Identify an extraneous solution to a radical equation
Marcy solved the radical equation square root of, x, plus, 2, end square root, equals, x and got two solutions, 2 and minus, 1.
When we substitute 2 for x into the equation, the left side of the equation is square root of, 2, plus, 2, end square root, equals
and the right side of the equation is 2. Therefore, 2 is
.
When we substitute minus, 1 for x into the equation, the left side of the equation is square root of, minus, 1, plus, 2, end square root, equals
and the right side of the equation is minus, 1. Therefore, minus, 1 is
.

## How do I solve rational equations?

### Equations with rational expressions

Equations with rational expressionsSee video transcript

### What do I need to know to solve rational equations?

Knowledge of fractions, polynomial operations and factoring, and quadratic equations is essential for successfully solving rational equations.
To solve a rational equation:
1. Rewrite the equation until the variable no longer appears in the denominators of rational expressions.
2. Rearrange and solve the resulting linear or quadratic equation.

Example: If start fraction, 1, divided by, x, plus, 1, end fraction, equals, start fraction, 2, divided by, x, end fraction, what is the value of x ?

Most often, the reason a solution to a rational equation is extraneous is because the solution, when substituted into the original equation, results in division by 0. For example, if one of the solutions to a rational equation is 2 and the original equation contains the denominator x, minus, 2, then the solution 2 is extraneous because 2, minus, 2, equals, 0, and we cannot divide by 0.
To check for extraneous solutions to a rational equation:
1. Solve the rational equation as outlined above.
2. Substitute the solution(s) into the original equation. A solution is extraneous if it does not satisfy the original equation.

Example: What value(s) of x satisfies the equation start fraction, 2, divided by, x, minus, 1, end fraction, equals, start fraction, x, plus, 1, divided by, x, minus, 1, end fraction ?

### Try it!

TRY: Identify the steps to solving a rational equation
start fraction, 3, x, divided by, x, plus, 2, end fraction, equals, 2
To solve the equation above, we first
both sides of the equation by x, plus, 2, then solve the resulting
equation.
Because the denominator of the rational expression is x, plus, 2, the only value of x that would lead to division by 0 is
. Therefore, when we get 4 as the solution, we know that it is
.

TRY: Identify an extraneous solution to a rational equation
Mehdi solved the rational equation start fraction, x, squared, minus, 4, divided by, x, plus, 2, end fraction, equals, 4 and got two solutions, minus, 2 and 6.
When we substitute minus, 2 for x into the equation, the denominator of the rational expression is minus, 2, plus, 2, equals
. Therefore, minus, 2 is
.
When we substitute 6 for x into the equation, the denominator of the rational expression is 6, plus, 2, equals
and the rational expression is equal to start fraction, 6, squared, minus, 4, divided by, 6, plus, 2, end fraction, equals
. Therefore, 6 is
.

square root of, 6, x, plus, 9, end square root, equals, x, plus, 3
Which of the following values of x satisfies the equation above?

Practice: check for extraneous solutions to a radical equation
square root of, 4, x, plus, 16, end square root, equals, x, plus, 1
Which of the following are the solutions to the equation above?
start text, I, end text. minus, 3
start text, I, I, end text. 5

Practice: solve a rational equation
If start fraction, 1, divided by, x, plus, 1, end fraction, equals, start fraction, 3, divided by, 5, x, minus, 1, end fraction, what is the value x ?

Practice: solve a rational equation
Which of the following values of x satisfies the equation start fraction, x, squared, minus, 4, x, divided by, x, minus, 4, end fraction, equals, 1 ?

## Things to remember

The radical operator (square root of, empty space, end square root) calculates only the positive square root. If a solution leads to equating the square root of a number to a negative number, then that solution is extraneous.
We cannot divide by 0. If a solution leads to division by 0, then that solution is extraneous.

## Want to join the conversation?

• Do you always have to use foil when squaring binomials and monomials?
• When you square a monomial, the exponent distributes to every factor in the monomial. (2x^2)^2 = 4x^4, not 2x^4.
When you square a binomial, you can use foil, but you can use a direct formula that's very similar to foiling it, just a bit faster.
(a + b)^2 = a^2 + 2ab + b^2
If we take (x - 8) as an example:
(x - 8)^2 = x^2 + 2*x*8 + 8^2
= x^2 + 16x + 64
• What if I have a radical in a rational equation, say the radical is in the denominator. Is there any lessons on solving equations like that?
• I'm not sure if there are any on Khan Academy, since it's a pretty specific topic. It's nice to learn something with an example, so let's say we have the equation 5 / cbrt(x) = 6x / 4. We know that to solve a rational equation, we have to multiply the variable out of the denominator, and that to solve a radical equation, we have to cancel the radical by raising both sides to the appropriate power. All we have to do to solve a rational equation with a radical then is to combine the two:
5 / cbrt(x) = 6x / 4
5 * 4 = 6x * cbrt(x)
20^3 = (6x)^3 * cbrt(x)^3
8000 = 216x^3 * x = 216 x^4
(8000 / 216)^(1/4) = x, which is about +/- 2.47
It's the same thing you do to solve any equation. Isolate the variables, and then simplify it down.
• How often would we need to check for extraneous solutions in the SAT and how would we know if we should? Also, do we only need to check for extraneous solutions when solving radical and rational equations?
• In one of the questions when I plug in "-3" for "x" to check if it is extranious the resulting equation in the explaination says that:

"root(4) =/= -2"

however root 4 could be "2" or "-2". Is this a mistake in the lesson? or is it not the right answer unless somehow both answers are present?
• This might be too late, but hopefully it helps.

While its true that both 2^2 and -2^2 equal 4, the square root operator can only give positive solutions. They explain it in this article, under the heading *What do I need to know to solve radical equations?*
(1 vote)
• Shouldn't 4^2 - 4 x 4 / 4 - 4 = 1 in the last question, because it makes the numerator 0 too. 0/0 = 1
(1 vote)
• That would work, except 0/0 doesn't really equal 1. 0/0 is undefined, and probably the easiest way to understand that is if you put it into an equation.
If we have 0/0 = x, we can rewrite that as 0x = 0. Now here, you can put in anything for x and still get 0 back out. This means that there is no answer to 0/0, which means that any value of x that makes the value 0/0 isn't a solution (or dividing by 0 at all, for that matter).
• What is the solution for √10-4x = √2x+3?
• Hey Danny
First, you would square both sides.
This leaves you with:
10-4x = 2x+3
Then you get the variables on one side and the constants on the other.
So you get:
10-3 = 2x+4x
Which is:
7 = 6x
Then divide both sides by 6 and you get:
7/6 = x
• Are radical and rational equations the only ones with extraneous solutions? Or is there more?
(1 vote)
• how do you solve 4/n+1 - 1/ n^2+7n+6 = 3/r-1
(1 vote)
• It looks like you have two variables, so you’d need to have another equation in order to get a number value for n or r, if that’s what you’re trying to do.
(1 vote)
• How do you solve problems like √ 10-4x = √ 2x+3?