Radical, rational, and absolute value equations | Lesson (article)

What are radical, rational, and absolute value equations?

Radical equations are equations in which variables appear under radical symbols ( $\sqrt{}$ ‍).

$\sqrt{2 x - 1} = x$ ‍ is a radical equation.

Rational equations are equations in which variables can be found in the denominators of rational expressions.

$\frac{1}{x + 1} = \frac{2}{x}$ ‍ 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.

Absolute value equations are equations in which variables appear within vertical bars ( $| |$ ‍).

$| x + 1 | = 2$ ‍ is an absolute value equation.

In this lesson, we'll learn to:

Solve radical and rational equations
Identify extraneous solutions to radical and rational equations
Solve absolute value equations

You can learn anything. Let's do this!

How do I solve radical equations?

Intro to square-root equations & extraneous solutions

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Intro to square-root equations & extraneous solutions

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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 $\sqrt{2 x - 1} = x$ ‍, what is the value of $x$ ‍ ?

$\begin{aligned} \sqrt{2 x - 1} & = x \\ {(\sqrt{2 x - 1})}^{2} & = x^{2} \\ 2 x - 1 & = x^{2} \\ 2 x - 1 - (2 x - 1) & = x^{2} - (2 x - 1) \\ 0 & = x^{2} - 2 x + 1 \end{aligned}$ ‍

We can factor $x^{2} - 2 x + 1$ ‍ into $(x + a) (x + b)$ ‍:

$a + b = - 2$ ‍
$a b = 1$ ‍

$a = - 1$ ‍ and $b = - 1$ ‍ would work:

$- 1 + (- 1) = - 2$ ‍
$(- 1) (- 1) = 1$ ‍

Now, we can solve for $x$ ‍:

$\begin{aligned} 0 & = x^{2} - 2 x + 1 \\ 0 & = (x - 1) (x - 1) \\ 0 & = x - 1 \\ 0 + 1 & = x - 1 + 1 \\ 1 & = x \end{aligned}$ ‍

$1$ ‍ 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, $\sqrt{4} = 2$ ‍, not both $- 2$ ‍ and $2$ ‍ even though $(- 2)^{2} = 2^{2} = 4$ ‍. If a solution leads to equating the square root of a number to a negative number, then that solution is extraneous.

In most cases, solving radical equations on the SAT involves squaring both sides of the radical equation. Raising both sides of an equation to an even power is not a reversible operation. For example, if $a$ ‍ is equal to $b$ ‍, then $a^{2}$ ‍ must be equal to $b^{2}$ ‍. However, if $a^{2}$ ‍ is equal to $b^{2}$ ‍, $a$ ‍ does not necessarily equal to $b$ ‍.

Let's look at a numerical example. For $a = - 2$ ‍ and $b = 2$ ‍, their squares are equal as $(- 2)^{2} = 2^{2} = 4$ ‍, but we know for a fact that $- 2 \neq 2$ ‍. Therefore, when we solve an equation whose solution steps include squaring both sides of the equation, we're solving $a^{2} = b^{2}$ ‍, and we must perform additional checks to make sure that $a$ ‍ is also equal to $b$ ‍.

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 $\sqrt{3 x + 4} = x$ ‍ ?

$\begin{aligned} \sqrt{3 x + 4} & = x \\ {(\sqrt{3 x + 4})}^{2} & = x^{2} \\ 3 x + 4 & = x^{2} \\ 2 x - 1 - (3 x + 4) & = x^{2} - (3 x + 4) \\ 0 & = x^{2} - 3 x - 4 \end{aligned}$ ‍

We can factor $x^{2} - 3 x - 4$ ‍ into $(x + a) (x + b)$ ‍:

$a + b = - 3$ ‍
$a b = - 4$ ‍

$a = - 4$ ‍ and $b = 1$ ‍ would work:

$- 4 + 1 = - 3$ ‍
$(- 4) (1) = - 4$ ‍

Now, we can solve for $x$ ‍:

$\begin{aligned} 0 & = x^{2} - 3 x - 4 \\ 0 & = (x - 4) (x + 1) \\ 0 & = x - 4 or 0 = x + 1 \\ 0 & = x - 4 \\ 0 + 4 & = x - 4 + 4 \\ 4 & = x \\ 0 & = x + 1 \\ 0 - 1 & = x + 1 - 1 \\ - 1 & = x \end{aligned}$ ‍

Now, we need to substitute $4$ ‍ and $- 1$ ‍ for $x$ ‍ in the original equation to check whether they're extraneous:

$\begin{aligned} \sqrt{3 (4) + 4} & \overset{?}{=} 4 \\ \sqrt{16} & \overset{?}{=} 4 \\ 4 & \overset{✓}{=} 4 \\ \sqrt{3 (- 1) + 4} & \overset{?}{=} - 1 \\ \sqrt{1} & \overset{?}{=} - 1 \\ 1 & \neq - 1 \end{aligned}$ ‍

$x = 4$ ‍ satisfies the original equation, but $x = - 1$ ‍ does not. As such, $x = - 1$ ‍ is an extraneous solution.

$4$ ‍ is the solution to the equation $\sqrt{3 x + 4} = x$ ‍.

Try it!

Try: identify the steps to solving a radical equation

$\sqrt{2 x - 9} = x - 6$ ‍

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.

How do I solve rational equations?

Equations with rational expressions

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Equations with rational expressions

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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 $\frac{1}{x + 1} = \frac{2}{x}$ ‍, what is the value of $x$ ‍ ?

$\begin{aligned} \frac{1}{x + 1} & = \frac{2}{x} \\ \frac{1}{x + 1} \cdot (x + 1) & = \frac{2}{x} \cdot (x + 1) \\ 1 & = \frac{2 (x + 1)}{x} \\ 1 \cdot x & = \frac{2 (x + 1)}{x} \cdot x \\ x & = 2 (x + 1) \\ x & = 2 x + 2 \\ x - x & = 2 x + 2 - x \\ 0 & = x + 2 \\ 0 - 2 & = x + 2 - 2 \\ - 2 & = x \end{aligned}$ ‍

$- 2$ ‍ 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 $x - 2$ ‍, then the solution $2$ ‍ is extraneous because $2 - 2 = 0$ ‍, 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 $\frac{2}{x - 1} = \frac{x + 1}{x - 1}$ ‍ ?

$\begin{aligned} \frac{2}{x - 1} & = \frac{x + 1}{x - 1} \\ \frac{2}{x - 1} \cdot (x - 1) & = \frac{x + 1}{x - 1} \cdot (x - 1) \\ 2 & = x + 1 \\ 2 - 1 & = x + 1 - 1 \\ 1 & = x \end{aligned}$ ‍

However, when we substitute $x = 1$ ‍ into the original equation, the denominator of both sides of the equation becomes $1 - 1 = 0$ ‍. Since we cannot divide by $0$ ‍, $1$ ‍ is an extraneous solution.

No value of $x$ ‍ satisfies the equation $\frac{2}{x - 1} = \frac{x + 1}{x - 1}$ ‍.

Try it!

TRY: Identify the steps to solving a rational equation

$\frac{3 x}{x + 2} = 2$ ‍

To solve the equation above, we first

both sides of the equation by $x + 2$ ‍, then solve the resulting

equation.

Because the denominator of the rational expression is $x + 2$ ‍, 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 $\frac{x^{2} - 4}{x + 2} = 4$ ‍ and got two solutions, $- 2$ ‍ and $6$ ‍.

When we substitute $- 2$ ‍ for $x$ ‍ into the equation, the denominator of the rational expression is $- 2 + 2 =$ ‍

. Therefore, $- 2$ ‍ is

When we substitute $6$ ‍ for $x$ ‍ into the equation, the denominator of the rational expression is $6 + 2 =$ ‍

and the rational expression is equal to $\frac{6^{2} - 4}{6 + 2} =$ ‍

. Therefore, $6$ ‍ is

How do I solve absolute value equations?

Absolute value equation with two solutions

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Worked example: absolute value equation with two solutions

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Absolute value equation with no solution

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Worked example: absolute value equations with no solution

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The absolute value of a number is equal to the number's distance from $0$ ‍ on the number line, which means the absolute value of a nonzero number is always positive. For example:

The absolute value of $2$ ‍, or $| 2 |$ ‍, is $2$ ‍.
The absolute value of $- 2$ ‍, or $| - 2 |$ ‍, is also $2$ ‍.

Practically, this means every absolute value equation can be split into two linear equations. For example, if $| 2 x + 1 | = 5$ ‍:

The absolute value equation is true if $2 x + 1 = 5$ ‍.
The absolute value equation is also true if $2 x + 1 = - 5$ ‍ since $| - 5 | = 5$ ‍.

When solving absolute value equations, rewrite the equation as two linear equations, then solve each linear equation. Both solutions are solutions to the absolute value equation.

Example: What are the solutions to the equation $| 2 x - 1 | = 5$ ‍ ?

The absolute value equation can be divided into two linear equations:

$\begin{aligned} 2 x - 1 & = 5 & first equation \\ 2 x - 1 + 1 & = 5 + 1 \\ 2 x & = 6 \\ \frac{2 x}{2} & = \frac{6}{2} \\ x & = 3 & first solution \\ 2 x - 1 & = - 5 & second equation \\ 2 x - 1 + 1 & = - 5 + 1 \\ 2 x & = - 4 \\ \frac{2 x}{2} & = \frac{- 4}{2} \\ x & = - 2 & second solution \end{aligned}$ ‍

The solutions are $3$ ‍ and $- 2$ ‍.

Try it!

try: write two linear equations from an absolute value equation

$| 3 x + 7 | = 16$ ‍

To solve the absolute value equation above, we must solve two linear equations.

$x = 3$ ‍ is one solution to the absolute value equation and satisfies the linear equation

$x = - \frac{23}{3}$ ‍ is the other solution to the absolute value equation and satisfies the linear equation

Your turn!

Practice: solve a radical equation

$\sqrt{6 x + 9} = x + 3$ ‍

Which of the following values of $x$ ‍ satisfies the equation above?

Choose 1 answer:

(Choice A)
$0$ ‍
(Choice B)
$3$ ‍
(Choice C)
$6$ ‍
(Choice D)
$9$ ‍

Practice: check for extraneous solutions to a radical equation

$\sqrt{4 x + 16} = x + 1$ ‍

Which of the following are the solutions to the equation above?

$I$ ‍. $- 3$ ‍
$II$ ‍. $5$ ‍

Choose 1 answer:

(Choice A)
$I$ ‍ only
(Choice B)
$II$ ‍ only
(Choice C)
Both $I$ ‍ and $II$ ‍
(Choice D)
Neither $I$ ‍ nor $II$ ‍

Practice: solve a rational equation

If $\frac{1}{x + 1} = \frac{3}{5 x - 1}$ ‍, what is the value $x$ ‍ ?

Choose 1 answer:

(Choice A)
$\frac{1}{3}$ ‍
(Choice B)
$\frac{1}{2}$ ‍
(Choice C)
$1$ ‍
(Choice D)
$2$ ‍

Practice: solve a rational equation

Which of the following values of $x$ ‍ satisfies the equation $\frac{x^{2} - 4 x}{x - 4} = 1$ ‍ ?

Choose 1 answer:

(Choice A)
$1$ ‍ only
(Choice B)
$4$ ‍ only
(Choice C)
$1$ ‍ and $4$ ‍
(Choice D)
No such value of $x$ ‍ exists.

practice: solve an absolute value equation

$| x - 7 | = 1$ ‍

If $a$ ‍ and $b$ ‍ are solutions to the equation above, what is the value of $a + b$ ‍ ?

Things to remember

The radical operator ( $\sqrt{}$ ‍) 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.

For the absolute value equation $| a x + b | = c$ ‍, rewrite the equation as the following linear equations and solve them.

$a x + b = c$ ‍
$a x + b = - c$ ‍

Both solutions are solutions to the absolute value equation.

Radical, rational, and absolute value equations | Lesson (article) | Khan Academy (2024)

What are radical, rational, and absolute value equations?

How do I solve radical equations?

Intro to square-root equations & extraneous solutions

What do I need to know to solve radical equations?

Try it!

How do I solve rational equations?

Equations with rational expressions

What do I need to know to solve rational equations?

Try it!

How do I solve absolute value equations?

Absolute value equation with two solutions

Absolute value equation with no solution

Try it!

Your turn!

Things to remember

References