2.6: Other Types of Equations
- Solve equations using factoring.
- Solve radical equations.
- Solve absolute value equations.
- Solve other types of equations.
Solving Other Types of Equations
We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at radical equations, rational equations and absolute value equations. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.
Solving Radical Equations
Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as
\[\sqrt{3x+18}=x \nonumber\]
\[\sqrt{x+3}=x-3 \nonumber\]
\[\sqrt{x+5}-\sqrt{x-3}=2 \nonumber\]
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.
An equation containing variable terms in the radicand is called a radical equation.
- 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 \(n^{th}\) root radical, raise both sides to the \(n^{th}\) power. Doing so eliminates the radical symbol.
- Solve the remaining equation.
- If a radical term still remains, repeat steps 1–2.
- Confirm solutions by substituting them into the original equation.
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.
\[\begin{align*} \sqrt{15-2x}&= x\\ {\left (\sqrt{15-2x} \right )}^2&= {(x)}^2\\ 15-2x&= x^2 \end{align*}\]
We see that the remaining equation is a quadratic. Set it equal to zero and solve.
\[\begin{align*} 0&= x^2+2x-15\\ 0&= (x+5)(x-3)\\ x&= -5\\ x&= 3 \end{align*}\]
The proposed solutions are \(−5\) and \(3\). Let us check each solution back in the original equation. First, check \(x=−5\).
\[\begin{align*} \sqrt{15-2x}&= x\\ \sqrt{15-2(-5)}&=-5\\ \sqrt{25}&= -5\\ 5&\neq -5 \end{align*}\]
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 \(x=3\).
\[\begin{align*} \sqrt{15-2x}&= x\\ \sqrt{15-2(3)}&= 3\\ \sqrt{9}&= 3\\ 3&= 3 \end{align*}\]
The solution is \(3\).
Solve the radical equation: \(\sqrt{x+3}=3x-1\)
- Answer
-
\(x=1\), extraneous solution \(x=−\dfrac{2}{9}\)
Solve \(\sqrt{2x+3}+\sqrt{x-2}=4\)
Solution
As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.
\[\sqrt{2x+3}+\sqrt{x-2}=4 \nonumber\]
\[\begin{align*} \sqrt{2x+3}&= 4-\sqrt{x-2} \qquad \text{Subtract } \sqrt{x-2} \text{ from both sides}\\ {\left (\sqrt{2x+3} \right )}^2&= {\left (4-\sqrt{x-2} \right )}^2\qquad \text{Square both sides} \end{align*}\]
Use the perfect square formula to expand the right side: \({(a−b)}^2=a^2−2ab+b^2\).
\[\begin{align*} 2x+3&= {(4)}^2-2(4)\sqrt{x-2}+{(\sqrt{x-2})}^2\\ 2x+3&= 16-8\sqrt{x-2}+(x-2)\\ 2x+3&= 14+x-8\sqrt{x-2} \qquad \text{Combine like terms}\\ x-11&= -8\sqrt{x-2} \qquad \text{Isolate the second radical}\\ {(x-11)}^2&= {(-8\sqrt{x-2})}^2 \qquad \text{Square both sides}\\ x^2-22x+121&= 64(x-2) \end{align*}\]
Now that both radicals have been eliminated, set the quadratic equal to zero and solve.
\[\begin{align*} x^2-22x+121&= 64x-128\\ x^2-86x+249&= 0\\ (x-3)(x-83)&= 0\\ x&= 3\\ x&= 83 \end{align*}\]
The proposed solutions are \(3\) and \(83\). Check each solution in the original equation.
\[\begin{align*} \sqrt{2x+3}+\sqrt{x-2}&= 4\\ \sqrt{2x+3}&= 4-\sqrt{x-2}\\ \sqrt{2(3)+3}&= 4-\sqrt{(3)-2}\\ \sqrt{9}&= 4-\sqrt{1}\\ 3&= 3 \end{align*}\]
One solution is \(3\).
Check \(x=83\).
\[\begin{align*} \sqrt{2x+3}+\sqrt{x-2}&= 4\\ \sqrt{2x+3}&= 4-\sqrt{x-2}\\ \sqrt{2(83)+3}&= 4-\sqrt{(83)-2}\\ \sqrt{169}&= 4-\sqrt{81}\\ 13&\neq -5 \end{align*}\]
The only solution is \(3\). We see that \(x=83\) is an extraneous solution.
Solve the equation with two radicals: \(\sqrt{3x+7}+\sqrt{x+2}=1\)
- Answer
-
\(x=−2\), extraneous solution \(x=−1\)
Solving a Rational Equation
A r ational equation contains at least one rational expression where the variable appears in at least one of the denominators.
In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation . Recall that a rational number is the ratio of two numbers, such as \(\dfrac{2}{3}\) or \(\dfrac{7}{2}\). A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.
\[\dfrac{x+1}{x^2-4} \nonumber \]
\[\dfrac{1}{x-3} \nonumber \]
or
\[\dfrac{4}{x^2+x-2} \nonumber \]
Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD). Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.
Solve the rational equation:
\[\dfrac{7}{2x}-\dfrac{5}{3x}=\dfrac{22}{3} \nonumber \]
Solution
We have three denominators; \(2x\),\(3x\), and \(3\). The LCD must contain \(2x\),\(3x\), and \(3\). An LCD of \(6x\) contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD \(6x\).
\[\begin{align*}
(6x)\left[\dfrac{7}{2x}-\dfrac{5}{3x}\right]&=\left[\dfrac{22}{3}\right](6x)\\
(6x)\left(\dfrac{7}{2x}\right)-(6x)\left(\dfrac{5}{3x}\right)&=\left(\dfrac{22}{3}\right)(6x)\qquad \text{Use the distributive property. Cancel out the common factors}\\
3(7)-2(5)&=22(2x)\qquad \text{Multiply remaining factors by each numerator.}\\
21-10&=44x\\
11&=44x\\
\dfrac{11}{44}&=x\\
\dfrac{1}{4}&=x
\end{align*}\]
A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as \((x+1)\). Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are \(x\), \(x−1\), and \(3x−3\). First, factor all denominators. We then have \(x\), \((x−1)\), and \(3(x−1)\) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of \((x−1)\). The x in the first denominator is separate from the \(x\) in the \((x−1)\) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the \(x\), one factor of \((x−1)\), and the 3. Thus, the LCD is the following:
\(x(x−1)3=3x(x−1)\)
So, both sides of the equation would be multiplied by \(3x(x−1)\). Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.
Another example is a problem with two denominators, such as \(x\) and \(x^2+2x\). Once the second denominator is factored as \(x^2+2x=x(x+2)\), there is a common factor of \(x\) in both denominators and the LCD is \(x(x+2)\).
Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.
\[\dfrac{a}{b}=\dfrac{c}{d}\]
We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.
Multiply a(d) and b(c), which results in \(ad=bc\).
Any value that makes a denominator in the original expression equal to zero must be excluded from the possible solutions. This value is called an exclusion.
- Factor all denominators in the equation.
- Find and exclude values that set each denominator equal to zero.
- Find the LCD.
- Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
- Solve the remaining equation.
- Make sure to plug in all possible solutions back into the original equation to avoid producing zero in a denominator
Solve the following rational equation:
\(\dfrac{2}{x}-\dfrac{3}{2}=\dfrac{7}{2x}\)
SolutionWe have three denominators: \(x\), \(2\), and \(2x\). No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is \(2x\). Only one value is excluded from a solution set, \(0\). Next, multiply the whole equation (both sides of the equal sign) by \(2x\).
\[\begin{align*} 2x\left[\dfrac{2}{x}-\dfrac{3}{2}\right]&=\left[\dfrac{7}{2x}\right](2x)\\ 2x\left(\dfrac{2}{x}\right)-2x\left(\dfrac{3}{2}\right)&=\left(\dfrac{7}{2x}\right)(2x)\qquad \text{Distribute } 2x\\ 2(2)-3x&=7\qquad \text{Denominators cancel out.}\\ 4-3x&=7\\ -3x&=3\\ x&=-1 \text { or } \{-1\} \end{align*}\]
The proposed solution is \(−1\), which is not an excluded value, so the solution set contains one number, \(x=−1\), or \(\{−1\}\) written in set notation.
Solve the rational equation:
\(\dfrac{2}{3x}=\dfrac{1}{4}-\dfrac{1}{6x}\)
- Answer
-
\(x=\dfrac{10}{3}\)
Solve the following rational equation:
\(\dfrac{1}{x}=\dfrac{1}{10}-\dfrac{3}{4x}\)
Solution
First find the common denominator. The three denominators in factored form are \(x,10=2⋅5\), and \(4x=2⋅2⋅x\). The smallest expression that is divisible by each one of the denominators is \(20x\). Only \(x=0\) is an excluded value. Multiply the whole equation by \(20x\).
\[\begin{align*} 20x\left(\dfrac{1}{x}\right)&= \left(\dfrac{1}{10}-\dfrac{3}{4x}\right)20x\\ 20&= 2x-15\\ 35&= 2x\\ \dfrac{35}{2}&= x \end{align*}\]
The solution is \(\dfrac{35}{2}\).
Solve the rational equation:
\[-\dfrac{5}{2x}+\dfrac{3}{4x}=-\dfrac{7}{4} \nonumber \]
- Answer
-
\(x=1\)
Solve the following rational equations and state the excluded values:
- \(\dfrac{3}{x-6}=\dfrac{5}{x}\)
- \(\dfrac{x}{x-3}=\dfrac{5}{x-3}-\dfrac{1}{2}\)
- \(\dfrac{x}{x-2}=\dfrac{5}{x-2}-\dfrac{1}{2}\)
Solution
a.
The denominators \(x\) and \(x−6\) have nothing in common. Therefore, the LCD is the product \(x(x−6)\). However, for this problem, we can cross-multiply.
\[\begin{align*} \dfrac{3}{x-6}&=\dfrac{5}{x}\\ 3x&=5(x-6)\qquad \text{Distribute.}\\ 3x&=5x-30\\ -2x&=-30\\ x&=15 \end{align*}\]
The solution is \(15\). The excluded values are \(6\) and \(0\).
b.
The LCD is \(2(x−3)\). Multiply both sides of the equation by \(2(x−3)\).
\[\begin{align*} 2(x-3)\left [\dfrac{x}{x-3} \right ]&= \left [\dfrac{5}{x-3}-\dfrac{1}{2} \right ]2(x-3)\\ \dfrac{2(x-3)x}{x-3}&= \dfrac{2(x-3)5}{x-3}-\dfrac{2(x-3)}{2}\\ 2x&= 10-(x-3)\\ 2x&= 13-x\\ 3x&= 13\\ x&= \dfrac{13}{3} \end{align*}\]
The solution is \(\dfrac{13}{3}\). The excluded value is \(3\).
c.
The least common denominator is \(2(x−2)\). Multiply both sides of the equation by \(x(x−2)\).
\[\begin{align*} 2(x-2)\left [\dfrac{x}{x-2} \right ]&= \left [\dfrac{5}{x-2}-\dfrac{1}{2} \right ]2(x-2)\\ 2x&= 10-(x-2)\\ 2x&= 12-x\\ 3x&= 12\\ x&= 4 \end{align*}\]
The solution is \(4\). The excluded value is \(2\).
Solve \(\dfrac{-3}{2x+1}=\dfrac{4}{3x+1}\). State the excluded values.
- Answer
-
\(x=-\dfrac{7}{17}\). Excluded values are \(x=−12\) and \(x=−13\).
Solve the rational equation after factoring the denominators: \(\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{x^2-1}\). State the excluded values.
Solution
We must factor the denominator \(x^2−1\). We recognize this as the difference of squares, and factor it as \((x−1)(x+1)\). Thus, the LCD that contains each denominator is \((x−1)(x+1)\). Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.
\[\begin{align*} (x+1)(x-1)\left [\dfrac{2}{x+1}-\dfrac{1}{x-1} \right ]&= \left [\dfrac{2x}{x^2-1} \right ](x+1)(x-1)\\ 2(x-1)-(x+1)&= 2x\\ 2x-2-x-1&= 2x \text{ Distribute the negative sign}\\ -3-x&= 0\\ x&= -3 \end{align*}\]
The solution is \(−3\). The excluded values are \(1\) and \(−1\).
Solve the rational equation:
\(\dfrac{2}{x-2}+\dfrac{1}{x+1}=\dfrac{1}{x^2-x-2}\)
- Answer
-
\(x=\dfrac{1}{3}\)
Solving Rational Equations Resulting 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.
Solve the following rational equation: \(\dfrac{-4x}{x-1}+\dfrac{4}{x+1}=\dfrac{-8}{x^2-1}\)
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.
\[\begin{align*} (x+1)(x-1)\left (\dfrac{-4x}{x-1}+\dfrac{4}{x+1} \right )&= \left (\dfrac{-8}{x^2-1} \right )(x+1)(x-1)\\ -4x(x+1)+4(x-1)&= -8\\ -4x^2-4x+4x-4&= -8\\ -4x^2+4&= 0\\
-4(x^2-1)&= 0\\ -4(x+1)(x-1)&= 0\\ x&= -1\\ x&= 1 \end{align*}\]
In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.
Solve \(\dfrac{3x+2}{x-2}+\dfrac{1}{x}=\dfrac{-2}{x^2-2x}\)
- Answer
-
\(x=−1, x= 0\) is not a solution.
Understanding Absolute Value
Recall that in its basic form \(f(x)=|x|\), the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.
The absolute value can be defined as a piecewise equation
\[|x|= \begin{cases} x & \text{ if }x \text{ } {\geq} \text{ } 0 \\ -x & \text{ if } x<0 \end{cases}\]
Example \(\PageIndex{1}\): Determine a Number within a Prescribed Distance
Describe all values \(x\) within or including a distance of 4 from the number 5.
Solution
We want the distance between \(x\) and 5 to be less than or equal to 4. We can draw a number line, such as the one in , to represent the condition to be satisfied.
The distance from \(x\) to 5 can be represented using the absolute value as \(|x−5|\). We want the values of \(x\) that satisfy the condition \(| x−5 |\leq4\).
Analysis
Note that
\[\begin{align*} -4&{\leq}x-5 & x-5&\leq4 \\[4pt] 1&{\leq}x & x&{\leq}9 \end{align*}\]
So \(|x−5|\leq4\) is equivalent to \(1{\leq}x\leq9\).
However, mathematicians generally prefer absolute value notation.
Exercise \(\PageIndex{1}\)
Describe all values \(x\) within a distance of 3 from the number 2.
- Answer
-
\(|x−2|\leq3\)
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.
\[\begin{align*} 2x-6&= 8\\ 2x&= 14\\ x&= 7 \end{align*}\]
OR
\[\begin{align*} 2x-6&= -8\\ 2x&= -2\\ x&= -1 \end{align*}\]
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.
For real number expressions \(A\) and \(B\), an equation of the form \(|A|=B\), with \(B>0\), will have TWO solutions when \(A=B\) or \(A=−B\).
If \(B=0\), the equation \(|A|=B\) has ONE solution, \(A=0\).
If \(B<0\), the equation \(|A|=B\) has no solution.
An absolute value equation in the form \(|ax+b|=c\) has the following properties:
- If \(c<0\),\(|ax+b|=c\) has no solution.
- If \(c=0\),\(|ax+b|=c\) has one solution.
- If \(c>0\),\(|ax+b|=c\) has two solutions.
Given an absolute value equation, solve it.
- Isolate the absolute value expression on one side of the equal sign.
- If \(c>0\), write and solve two equations: \(ax+b=c\) and \(ax+b=−c\).
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:
\[\begin{align*} 6x+4&= 8\\ 6x&= 4\\ x&= \dfrac{2}{3} \end{align*}\]
OR
\[\begin{align*} 6x+4&= -8\\ 6x&= -12\\ x&= -2 \end{align*}\]
The two solutions are \(\dfrac{2}{3}\) 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.
\[\begin{align*} |3x-5|-4&= 6\\ |3x-5|&= 10\\ 3x-5&= 10\\ 3x&= 15\\ x&= 5 \end{align*}\]
OR
\[\begin{align*} 3x-5&= -10\\ 3x=-5\\ x=\dfrac{5}{3} \end{align*}\]
There are two solutions: \(5\), and \(-\dfrac{5}{3}\).
- \(|−5x+10|=0\)
The equation is set equal to zero, so we have to write only one equation.
\[\begin{align*} -5x+10&= 0\\ -5x&= -10\\ x&= 2 \end{align*}\]
There is one solution: \(2\ ).
Solve the absolute value equation: \(|1−4x|+8=13\).
- Answer
-
\(x=−1, x=\dfrac{3}{2}\)
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 Example , Example , and Example .
- Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See Example and Example .
- We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See Example and Example .
- To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See Example .
- 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 Example and Example .
- Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See Example .