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Other Types of Equations

Solve equations involving rational exponents.

Solving Equations Involving Rational Exponents

Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, \({16}^{\frac{1}{2}}\) is another way of writing \(\sqrt{16};\) \({8}^{\frac{1}{3}}\) is another way of writing \(\text{}\ \sqrt[3]{8}.\) The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.

We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, \(\frac{2}{3}(\frac{3}{2})=1,\) \(3(\frac{1}{3})=1,\) and so on.

Example

Try it.

Evaluate \({8}^{\frac{2}{3}}.\)

Solution

Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite \({8}^{\frac{2}{3}}\) as \({({8}^{\frac{1}{3}})}^{2}.\)

\[\begin{array}{lll}{({8}^{\frac{1}{3}})}^{2} & = & {(2)}^{2} \\ & = & 4\end{array}\]

Example

Try it.

Solve the equation in which a variable is raised to a rational exponent: \({x}^{\frac{5}{4}}=32.\)

Solution

The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of \(\frac{5}{4},\) which is \(\frac{4}{5}.\)

\[\begin{array}{llll}{x}^{\frac{5}{4}} & = & 32 & \\ {({x}^{\frac{5}{4}})}^{\frac{4}{5}} & = & {(32)}^{\frac{4}{5}} & \\ x & = & {(2)}^{4} & \ \text{The fifth root of 32 is 2.} \\ & = & 16 & \end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Equations Using Factoring

We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.

Example

Try it.

Solve the polynomial by factoring: \(5{x}^{4}=80{x}^{2}.\)

Solution

First, set the equation equal to zero. Then factor out what is common to both terms, the GCF.

\[\begin{array}{lll}5{x}^{4}-80{x}^{2} & = & 0 \\ 5{x}^{2}({x}^{2}-16) & = & 0\end{array}\]

Notice that we have the difference of squares in the factor \({x}^{2}-16,\) which we will continue to factor and obtain two solutions. The first term, \(5{x}^{2},\) generates, technically, two solutions as the exponent is 2, but they are the same solution.

\[\begin{array}{lll}5{x}^{2} & = & 0 \\ x & = & 0 \\ {x}^{2}-16 & = & 0 \\ (x-4)(x+4) & = & 0 \\ x & = & 4 \\ x & = & -4\end{array}\]

The solutions are \(0\ \text{(double solution),}\) \(4,\) and \(-4.\)

Example

Try it.

Solve a polynomial by grouping: \({x}^{3}+{x}^{2}-9x-9=0.\)

Solution

This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.

\[\begin{array}{lll}{x}^{3}+{x}^{2}-9x-9 & = & 0 \\ {x}^{2}(x+1)-9(x+1) & = & 0 \\ ({x}^{2}-9)(x+1) & = & 0\end{array}\]

The grouping process ends here, as we can factor \({x}^{2}-9\) using the difference of squares formula.

\[\begin{array}{lll}({x}^{2}-9)(x+1) & = & 0 \\ (x-3)(x+3)(x+1) & = & 0 \\ x & = & 3 \\ x & = & -3 \\ x & = & -1\end{array}\]

The solutions are \(3,\) \(-3,\) and \(-1.\) Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in .

Solving Radical Equations

Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as

\[\begin{array}{lll}\sqrt{3x+18} & = & x \\ \sqrt{x+3} & = & x-3 \\ \sqrt{x+5}-\sqrt{x-3} & = & 2\end{array}\]

Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.

Example

Try it.

Solve \(\sqrt{15-2x}=x.\)

Solution

The radical is already isolated on the left side of the equal side, so proceed to square both sides.

\[\begin{array}{lll}\sqrt{15-2x} & = & x \\ {(\sqrt{15-2x})}^{2} & = & {(x)}^{2} \\ 15-2x & = & {x}^{2}\end{array}\]

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

\[\begin{array}{lll}0 & = & {x}^{2}+2x-15 \\ & = & (x+5)(x-3) \\ x & = & -5 \\ x & = & 3\end{array}\]

The proposed solutions are \(-5\) and \(3.\) Let us check each solution back in the original equation. First, check \(x=-5.\)

\[\begin{array}{lll}\sqrt{15-2x} & = & x \\ \sqrt{15-2(-5)} & = & -5 \\ \sqrt{25} & = & -5 \\ 5 & \ne & -5\end{array}\]

This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.

Check \(x=3.\)

\[\begin{array}{lll}\sqrt{15-2x} & = & x \\ \sqrt{15-2(3)} & = & 3 \\ \sqrt{9} & = & 3 \\ 3 & = & 3\end{array}\]

The solution is \(3.\)

Condensed — the full section is in OpenStax College Algebra 2e.

Solving an Absolute Value Equation

Next, we will learn how to solve an absolute value equation. To solve an equation such as \(|2x-6|=8,\) we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is \(8\) or \(-8.\) This leads to two different equations we can solve independently.

\[\begin{array}{lllllll}2x-6 & = & 8 & \ \text{or}\ & 2x-6 & = & -8 \\ 2x & = & 14 & & 2x & = & -2 \\ x & = & 7 & & x & = & -1\end{array}\]

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

Example

Try it.

Solve the following absolute value equations:

ⓐ \(|6x+4|=8\)
ⓑ \(|3x+4|=-9\)
ⓒ \(|3x-5|-4=6\)
ⓓ \(|-5x+10|=0\)

Solution

ⓐ
\(|6x+4|=8\)

Write two equations and solve each:
\[\begin{array}{lllllll}6x+4 & = & 8 & \ & 6x+4 & = & -8 \\ 6x & = & 4 & \ & 6x & = & -12 \\ x & = & \frac{2}{3} & \ & x & = & -2\end{array}\]
The two solutions are \(\frac{2}{3}\) and \(-2.\)
ⓑ
\(|3x+4|=-9\)
There is no solution as an absolute value cannot be negative.
ⓒ
\(|3x-5|-4=6\)
Isolate the absolute value expression and then write two equations.
\[\begin{array}{lllllllll} & & & |3x-5|-4 & = & 6 & & & \\ & & & |3x-5| & = & 10 & & & \\ 3x-5 & = & 10 & & & & 3x-5 & = & -10 \\ 3x & = & 15 & & & & 3x & = & -5 \\ x & = & 5 & & & & x & = & -\frac{5}{3}\end{array}\]
There are two solutions: \(5,\) and \(-\frac{5}{3}.\)
ⓓ
(d) \(|-5x+10|=0\)
The equation is set equal to zero, so we have to write only one equation.
\[\begin{array}{lll}-5x+10 & = & 0 \\ -5x & = & -10 \\ x & = & 2\end{array}\]
There is one solution: \(2.\)

Solving Other Types of Equations

There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic.

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

Example

Try it.

Solve the following rational equation: \(\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.\)

Solution

We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, \({x}^{2}-1=(x+1)(x-1).\) Then, the LCD is \((x+1)(x-1).\) Next, we multiply the whole equation by the LCD.

\[\begin{array}{lll}(x+1)(x-1)[\frac{-4x}{x-1}+\frac{4}{x+1}] & = & [\frac{-8}{(x+1)(x-1)}](x+1)(x-1) \\ -4x(x+1)+4(x-1) & = & -8 \\ -4{x}^{2}-4x+4x-4 & = & -8 \\ -4{x}^{2}+4 & = & 0 \\ -4({x}^{2}-1) & = & 0 \\ -4(x+1)(x-1) & = & 0 \\ x & = & -1 \\ x & = & 1\end{array}\]

In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See , , and .
Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See and .
We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See and .
To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See .
Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See and .
Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Evaluate \({8}^{\frac{2}{3}}.\)

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Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite \({8}^{\frac{2}{3}}\) as \({({8}^{\frac{1}{3}})}^{2}.\)
\[\begin{array}{lll}{({8}^{\frac{1}{3}})}^{2} & = & {(2)}^{2} \\ & = & 4\end{array}\]
Evaluate \({64}^{-\frac{1}{3}}.\)

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\(\frac{1}{4}\)
Solve the equation in which a variable is raised to a rational exponent: \({x}^{\frac{5}{4}}=32.\)

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The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal of \(\frac{5}{4},\) which is \(\frac{4}{5}.\)
\[\begin{array}{llll}{x}^{\frac{5}{4}} & = & 32 & \\ {({x}^{\frac{5}{4}})}^{\frac{4}{5}} & = & {(32)}^{\frac{4}{5}} & \\ x & = & {(2)}^{4} & \ \text{The fifth root of 32 is 2.} \\ & = & 16 & \end{array}\]
Solve the equation \({x}^{\frac{3}{2}}=125.\)

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\(25\)
Solve \(3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}.\)

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This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.
\[\begin{array}{lll}3{x}^{\frac{3}{4}}-({x}^{\frac{1}{2}}) & = & {x}^{\frac{1}{2}}-({x}^{\frac{1}{2}}) \\ 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}} & = & 0\end{array}\]
Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite \({x}^{\frac{1}{2}}\) as \({x}^{\frac{2}{4}}.\) Then, factor out \({x}^{\frac{2}{4}}\) from both terms on the left.
\[\begin{array}{lll}3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}} & = & 0 \\ {x}^{\frac{2}{4}}(3{x}^{\frac{1}{4}}-1) & = & 0\end{array}\]
Where did \({x}^{\frac{1}{4}}\) come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply \({x}^{\frac{2}{4}}\) back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to \(\frac{2}{4}\) equals \(\frac{3}{4}.\) Thus, the exponent on x in the parentheses is \(\frac{1}{4}.\)
Let us continue. Now we have two factors and can use the zero factor theorem.
\[\begin{array}{llll}{x}^{\frac{2}{4}}(3{x}^{\frac{1}{4}}-1) & = & 0 & \\ {x}^{\frac{2}{4}} & = & 0 & \\ x & = & 0 & \\ 3{x}^{\frac{1}{4}}-1 & = & 0 & \\ 3{x}^{\frac{1}{4}} & = & 1 & \\ {x}^{\frac{1}{4}} & = & \frac{1}{3} & \ \text{Divide both sides by 3}. \\ {({x}^{\frac{1}{4}})}^{4} & = & {(\frac{1}{3})}^{4} & \ \text{Raise both sides to the reciprocal of }\frac{1}{4}. \\ x & = & \frac{1}{81} & \end{array}\]
The two solutions are \(0\) and \(\frac{1}{81}.\)
Solve: \({(x+5)}^{\frac{3}{2}}=8.\)

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\(\{-1\}\)
Solve the polynomial by factoring: \(5{x}^{4}=80{x}^{2}.\)

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First, set the equation equal to zero. Then factor out what is common to both terms, the GCF.
\[\begin{array}{lll}5{x}^{4}-80{x}^{2} & = & 0 \\ 5{x}^{2}({x}^{2}-16) & = & 0\end{array}\]
Notice that we have the difference of squares in the factor \({x}^{2}-16,\) which we will continue to factor and obtain two solutions. The first term, \(5{x}^{2},\) generates, technically, two solutions as the exponent is 2, but they are the same solution.
\[\begin{array}{lll}5{x}^{2} & = & 0 \\ x & = & 0 \\ {x}^{2}-16 & = & 0 \\ (x-4)(x+4) & = & 0 \\ x & = & 4 \\ x & = & -4\end{array}\]
The solutions are \(0\ \text{(double solution),}\) \(4,\) and \(-4.\)
Solve by factoring: \(12{x}^{4}=3{x}^{2}.\)

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\(0,\) \(\frac{1}{2},\) \(-\frac{1}{2}\)
Solve a polynomial by grouping: \({x}^{3}+{x}^{2}-9x-9=0.\)

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This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.
\[\begin{array}{lll}{x}^{3}+{x}^{2}-9x-9 & = & 0 \\ {x}^{2}(x+1)-9(x+1) & = & 0 \\ ({x}^{2}-9)(x+1) & = & 0\end{array}\]
The grouping process ends here, as we can factor \({x}^{2}-9\) using the difference of squares formula.
\[\begin{array}{lll}({x}^{2}-9)(x+1) & = & 0 \\ (x-3)(x+3)(x+1) & = & 0 \\ x & = & 3 \\ x & = & -3 \\ x & = & -1\end{array}\]
The solutions are \(3,\) \(-3,\) and \(-1.\) Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in .
Solve \(\sqrt{15-2x}=x.\)

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The radical is already isolated on the left side of the equal side, so proceed to square both sides.
\[\begin{array}{lll}\sqrt{15-2x} & = & x \\ {(\sqrt{15-2x})}^{2} & = & {(x)}^{2} \\ 15-2x & = & {x}^{2}\end{array}\]
We see that the remaining equation is a quadratic. Set it equal to zero and solve.
\[\begin{array}{lll}0 & = & {x}^{2}+2x-15 \\ & = & (x+5)(x-3) \\ x & = & -5 \\ x & = & 3\end{array}\]
The proposed solutions are \(-5\) and \(3.\) Let us check each solution back in the original equation. First, check \(x=-5.\)
\[\begin{array}{lll}\sqrt{15-2x} & = & x \\ \sqrt{15-2(-5)} & = & -5 \\ \sqrt{25} & = & -5 \\ 5 & \ne & -5\end{array}\]
This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.
Check \(x=3.\)
\[\begin{array}{lll}\sqrt{15-2x} & = & x \\ \sqrt{15-2(3)} & = & 3 \\ \sqrt{9} & = & 3 \\ 3 & = & 3\end{array}\]
The solution is \(3.\)
Solve the radical equation: \(\sqrt{x+3}=3x-1\)

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\(1;\) extraneous solution \(-\frac{2}{9}\)
Solve \(\sqrt{2x+3}+\sqrt{x-2}=4.\)

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As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.
\[\begin{array}{llll}\sqrt{2x+3}+\sqrt{x-2} & = & 4 & \\ \sqrt{2x+3} & = & 4-\sqrt{x-2} & \ \text{Subtract }\sqrt{x-2}\ \text{from both sides}. \\ {(\sqrt{2x+3})}^{2} & = & {(4-\sqrt{x-2})}^{2} & \ \text{Square both sides}.\end{array}\]
Use the perfect square formula to expand the right side: \({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}.\)
\[\begin{array}{llll}2x+3 & = & {(4)}^{2}-2(4)\sqrt{x-2}+{(\sqrt{x-2})}^{2} & \\ 2x+3 & = & 16-8\sqrt{x-2}+(x-2) & \\ 2x+3 & = & 14+x-8\sqrt{x-2} & \ \text{Combine like terms}. \\ x-11 & = & -8\sqrt{x-2} & \ \text{Isolate the second radical}. \\ {(x-11)}^{2} & = & {(-8\sqrt{x-2})}^{2} & \ \text{Square both sides}. \\ {x}^{2}-22x+121 & = & 64(x-2) & \end{array}\]
Now that both radicals have been eliminated, set the quadratic equal to zero and solve.
\[\begin{array}{llll}{x}^{2}-22x+121 & = & 64x-128 & \\ {x}^{2}-86x+249 & = & 0 & \\ (x-3)(x-83) & = & 0 & \ \text{Factor and solve}. \\ x & = & 3 & \\ x & = & 83 & \end{array}\]
The proposed solutions are \(3\) and \(83.\) Check each solution in the original equation.
\[\begin{array}{lll}\sqrt{2x+3}+\sqrt{x-2} & = & 4 \\ \sqrt{2x+3} & = & 4-\sqrt{x-2} \\ \sqrt{2(3)+3} & = & 4-\sqrt{(3)-2} \\ \sqrt{9} & = & 4-\sqrt{1} \\ 3 & = & 3\end{array}\]
One solution is \(3.\)
Check \(x=83.\)
\[\begin{array}{lll}\sqrt{2x+3}+\sqrt{x-2} & = & 4 \\ \sqrt{2x+3} & = & 4-\sqrt{x-2} \\ \sqrt{2(83)+3} & = & 4-\sqrt{(83-2)} \\ \sqrt{169} & = & 4-\sqrt{81} \\ 13 & \ne & -5\end{array}\]
The only solution is \(3.\) We see that \(83\) is an extraneous solution.
Solve the equation with two radicals: \(\sqrt{3x+7}+\sqrt{x+2}=1.\)

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\(-2;\) extraneous solution \(-1\)
Solve the following absolute value equations:
1. ⓐ \(|6x+4|=8\)
2. ⓑ \(|3x+4|=-9\)
3. ⓒ \(|3x-5|-4=6\)
4. ⓓ \(|-5x+10|=0\)
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1. ⓐ
  \(|6x+4|=8\)
  
  Write two equations and solve each:
  \[\begin{array}{lllllll}6x+4 & = & 8 & \ & 6x+4 & = & -8 \\ 6x & = & 4 & \ & 6x & = & -12 \\ x & = & \frac{2}{3} & \ & x & = & -2\end{array}\]
  The two solutions are \(\frac{2}{3}\) and \(-2.\)
2. ⓑ
  \(|3x+4|=-9\)
  There is no solution as an absolute value cannot be negative.
3. ⓒ
  \(|3x-5|-4=6\)
  Isolate the absolute value expression and then write two equations.
  \[\begin{array}{lllllllll} & & & |3x-5|-4 & = & 6 & & & \\ & & & |3x-5| & = & 10 & & & \\ 3x-5 & = & 10 & & & & 3x-5 & = & -10 \\ 3x & = & 15 & & & & 3x & = & -5 \\ x & = & 5 & & & & x & = & -\frac{5}{3}\end{array}\]
  There are two solutions: \(5,\) and \(-\frac{5}{3}.\)
4. ⓓ
  (d) \(|-5x+10|=0\)
  The equation is set equal to zero, so we have to write only one equation.
  \[\begin{array}{lll}-5x+10 & = & 0 \\ -5x & = & -10 \\ x & = & 2\end{array}\]
  There is one solution: \(2.\)
Solve the absolute value equation: \(|1-4x|+8=13.\)

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\(-1,\) \(\frac{3}{2}\)
Solve this fourth-degree equation: \(3{x}^{4}-2{x}^{2}-1=0.\)

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This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let \(u={x}^{2}.\) Rewrite the equation in u.
\[3{u}^{2}-2u-1=0\]
Now solve the quadratic.
\[\begin{array}{lll}3{u}^{2}-2u-1 & = & 0 \\ (3u+1)(u-1) & = & 0\end{array}\]
Solve each factor and replace the original term for u.
\[\begin{array}{lll}3u+1 & = & 0 \\ 3u & = & -1 \\ u & = & -\frac{1}{3} \\ {x}^{2} & = & -\frac{1}{3} \\ x & = & \pm i\sqrt{\frac{1}{3}}\end{array}\]\[\begin{array}{lll}u-1 & = & 0 \\ u & = & 1 \\ {x}^{2} & = & 1 \\ x & = & \pm 1\end{array}\]
The solutions are \(\pm i\sqrt{\frac{1}{3}}\) and \(\pm 1.\)
Solve using substitution: \({x}^{4}-8{x}^{2}-9=0.\)

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\(-3,3,-i,i\)
Solve the equation in quadratic form: \({(x+2)}^{2}+11(x+2)-12=0.\)

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This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting \(u=x+2.\) Then rewrite the equation in u.
\[\begin{array}{lll}{u}^{2}+11u-12 & = & 0 \\ (u+12)(u-1) & = & 0\end{array}\]
Solve using the zero-factor property and then replace u with the original expression.
\[\begin{array}{lll}u+12 & = & 0 \\ u & = & -12 \\ x+2 & = & -12 \\ x & = & -14\end{array}\]
The second factor results in
\[\begin{array}{lll}u-1 & = & 0 \\ u & = & 1 \\ x+2 & = & 1 \\ x & = & -1\end{array}\]
We have two solutions: \(-14,\) and \(-1.\)
Solve: \({(x-5)}^{2}-4(x-5)-21=0.\)

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\(2,12\)
Solve the following rational equation: \(\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.\)

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We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, \({x}^{2}-1=(x+1)(x-1).\) Then, the LCD is \((x+1)(x-1).\) Next, we multiply the whole equation by the LCD.
\[\begin{array}{lll}(x+1)(x-1)[\frac{-4x}{x-1}+\frac{4}{x+1}] & = & [\frac{-8}{(x+1)(x-1)}](x+1)(x-1) \\ -4x(x+1)+4(x-1) & = & -8 \\ -4{x}^{2}-4x+4x-4 & = & -8 \\ -4{x}^{2}+4 & = & 0 \\ -4({x}^{2}-1) & = & 0 \\ -4(x+1)(x-1) & = & 0 \\ x & = & -1 \\ x & = & 1\end{array}\]
In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.
Solve \(\frac{3x+2}{x-2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}.\)

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\(-1\) \(\)(\(0\) is not a solution).
In a radical equation, what does it mean if a number is an extraneous solution?

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This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.
Explain why possible solutions must be checked in radical equations.
Your friend tries to calculate the value \(-{9}^{\frac{3}{2}}\) and keeps getting an ERROR message. What mistake are they probably making?

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They are probably trying to enter negative 9, but taking the square root of \(-9\) is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in \(-27.\)
Explain why \(|2x+5|=-7\) has no solutions.
Explain how to change a rational exponent into the correct radical expression.

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A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.
\({x}^{\frac{2}{3}}=16\)
\({x}^{\frac{3}{4}}=27\)

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\(x=81\)
\(2{x}^{\frac{1}{2}}-{x}^{\frac{1}{4}}=0\)
\({(x-1)}^{\frac{3}{4}}=8\)

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\(x=17\)
\({(x+1)}^{\frac{2}{3}}=4\)
\({x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}+6=0\)

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\(x=8,x=27\)
\({x}^{\frac{7}{3}}-3{x}^{\frac{4}{3}}-4{x}^{\frac{1}{3}}=0\)
\({x}^{3}+2{x}^{2}-x-2=0\)

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\(x=-2,1,-1\)
\(3{x}^{3}-6{x}^{2}-27x+54=0\)
\(4{y}^{3}-9y=0\)

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\(y=0,\frac{3}{2},\frac{-3}{2}\)
\({x}^{3}+3{x}^{2}-25x-75=0\)
\({m}^{3}+{m}^{2}-m-1=0\)

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\(m=1,-1\)
\(2{x}^{5}-14{x}^{3}=0\)
\(5{x}^{3}+45x=2{x}^{2}+18\)

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\(x=\frac{2}{5},\pm 3i\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Other Types of Equations

Solve equations involving rational exponents.
Solve equations using factoring.
Solve radical equations.
Solve absolute value equations.
Solve other types of equations.
Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an
Solve the remaining equation.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Попробуй сам.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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