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## SAT

### Course: SAT > Unit 6

Lesson 4: Passport to Advanced Math: lessons by skill- Solving quadratic equations | Lesson
- Interpreting nonlinear expressions | Lesson
- Quadratic and exponential word problems | Lesson
- Manipulating quadratic and exponential expressions | Lesson
- Radicals and rational exponents | Lesson
- Radical and rational equations | Lesson
- Operations with rational expressions | Lesson
- Operations with polynomials | Lesson
- Polynomial factors and graphs | Lesson
- Graphing quadratic functions | Lesson
- Graphing exponential functions | Lesson
- Linear and quadratic systems | Lesson
- Structure in expressions | Lesson
- Isolating quantities | Lesson
- Function Notation | Lesson

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# 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

### 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.

To solve a radical equation:

- Isolate the radical expression to one side of the equation.
- Square both sides the equation.
- 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:

- Solve the radical equation as outlined above.
- 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!

## How do I solve rational equations?

### Equations with rational expressions

### 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:

- Rewrite the equation until the variable no longer appears in the denominators of rational expressions.
- 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:

- Solve the rational equation as outlined above.
- 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!

## Your turn!

## 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?(5 votes)
- 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(4 votes)

- 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?(3 votes)
- 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.(3 votes)

- 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?(4 votes)
- 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?(2 votes)- 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).(3 votes)

- What is the solution for √10-4x = √2x+3?(0 votes)
- 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(4 votes)

- 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?(0 votes)
- Hi! Feeling a little confused by something not covered in this article. Why does cube root(a^8)= a^2cube root(a^2)? I know 8 is 2^3 but unsure why this expression doesn't just equal a^2.(0 votes)
- It's best if you think of this expression in terms of rational exponents. The cube root of anything is the same as raising it to the one-third, so the cube root of a to the eight can be written as a^(8/3). Now, all we do is split the fraction into a mixed number. 8/3 is the same as 2 2/3:

a^(8/3) = a^(2 + 2/3) = a^2 * a^(2/3)

= a^2 * cbrt(a^2)

Does this help?(3 votes)