Lesson 2.6: Other Types of Equations

New Concept

This lesson expands your problem-solving skills beyond linear and quadratic equations. You will learn to solve equations with rational exponents, radicals, absolute values, and those in quadratic form by applying specific algebraic techniques to isolate and find the variable.

What’s next

Next, you'll master these methods through a series of interactive examples and practice cards, starting with equations containing rational exponents.

Property

A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:

Examples

• Evaluate $27^{\frac{2}{3}}$. This can be rewritten as $(\sqrt[3]{27})^2$, which simplifies to $(3)^2 = 9$.
• Solve $x^{\frac{3}{2}} = 64$. Raise both sides to the $\frac{2}{3}$ power: $(x^{\frac{3}{2}})^{\frac{2}{3}} = 64^{\frac{2}{3}}$. This gives $x = (\sqrt[3]{64})^2 = 4^2 = 16$.
• Solve $5x^{\frac{3}{4}} - x^{\frac{1}{2}} = 0$. Factor out the term with the lowest exponent, $x^{\frac{1}{2}}$: $x^{\frac{1}{2}}(5x^{\frac{1}{4}} - 1) = 0$. This yields two solutions: $x=0$ and $x = (\frac{1}{5})^4 = \frac{1}{625}$.

Explanation

Fractional exponents combine powers and roots. The numerator is the power and the denominator is the root. To solve equations with them, raise both sides to the reciprocal power to isolate the variable, since a number times its reciprocal is 1.

Property

A polynomial of degree $n$ is an expression of the type

where $n$ is a positive integer and $a_n, \ldots, a_0$ are real numbers and $a_n \neq 0$. Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent $n$.

Examples

• Solve $3x^4 = 48x^2$. First, set the equation to zero: $3x^4 - 48x^2 = 0$. Factor out the GCF $3x^2$: $3x^2(x^2 - 16) = 0$. The solutions are $x=0$, and from $x^2-16=0$, $x=4$ and $x=-4$.
• Solve by grouping: $x^3 + 2x^2 - 16x - 32 = 0$. Group terms: $x^2(x+2) - 16(x+2) = 0$. Factor out $(x+2)$ to get $(x^2-16)(x+2)=0$. The solutions are $4, -4,$ and $-2$.
• Solve $2x^5 - 32x = 0$. Factor out the GCF $2x$: $2x(x^4 - 16) = 0$. Factor the difference of squares twice: $2x(x^2-4)(x^2+4)=0$, then $2x(x-2)(x+2)(x^2+4)=0$. The real solutions are $0, 2,$ and $-2$.

Explanation

For equations with powers higher than 2, you can often solve them by factoring. Move all terms to one side to equal zero, then factor using methods like GCF or grouping. Set each factor to zero to find the solutions.

Property

An equation containing terms with a variable in the radicand is called a radical equation.
To solve, first isolate the radical expression on one side.
Then, raise both sides to the power that matches the root (e.g., square both sides for a square root) to eliminate the radical.
Solve the resulting equation and always confirm solutions by substituting them back into the original equation to check for extraneous solutions.

Examples

• Solve $\sqrt{10 - 3x} = x$. Squaring both sides gives $10 - 3x = x^2$. Rearrange to $x^2 + 3x - 10 = 0$, which factors to $(x+5)(x-2)=0$. Checking reveals $x=-5$ is extraneous, so the only solution is $x=2$.
• Solve $\sqrt{x+7} = x-5$. Square both sides: $x+7 = x^2-10x+25$. Rearrange to $x^2-11x+18=0$, or $(x-2)(x-9)=0$. Checking shows $x=2$ is extraneous, leaving $x=9$ as the only solution.
• Solve $\sqrt{x+5} + \sqrt{x-3} = 4$. Isolate one radical: $\sqrt{x+5} = 4 - \sqrt{x-3}$. Square both sides: $x+5 = 16 - 8\sqrt{x-3} + (x-3)$. Simplify to $1 = \sqrt{x-3}$. Square again: $1 = x-3$, so $x=4$. Checking confirms this is a valid solution.

Explanation

To solve an equation with a variable under a radical, get the radical by itself and then raise both sides to a power to undo the root. This process can create false answers, so always plug your solutions back into the original equation to check!

Property

The absolute value of $x$ is written as $|x|$.
For real numbers $A$ and $B$, an equation of the form $|A| = B$, with $B \geq 0$, will have solutions when $A = B$ or $A = -B$.
If $B < 0$, the equation $|A| = B$ has no solution. To solve, first isolate the absolute value expression.
If the value it equals is positive, write two separate equations ($A=B$ and $A=-B$) and solve both.

Examples

• Solve $|2x - 1| = 9$. Set up two equations: $2x-1=9$ gives $x=5$, and $2x-1=-9$ gives $x=-4$. The solutions are $5$ and $-4$.
• Solve $|x+3| - 2 = 5$. First, isolate the absolute value: $|x+3|=7$. Set up two equations: $x+3=7$ gives $x=4$, and $x+3=-7$ gives $x=-10$.
• Solve $|4x+1| = -3$. An absolute value cannot equal a negative number. Therefore, this equation has no solution.

Explanation

Absolute value means distance from zero, which is always positive. So, $|A|=B$ means $A$ is either $B$ or $-B$. Isolate the absolute value part first, then create two separate equations to find the two possible solutions.

Property

If the exponent on the leading term is double the exponent on the middle term, we have an equation in quadratic form.
To solve, identify this pattern and substitute a variable (like $u$) for the variable part of the middle term.
Rewrite the equation as a standard quadratic, solve for $u$, then replace $u$ with the original term and solve for the original variable.

Examples

• Solve $x^4 - 7x^2 + 12 = 0$. Let $u=x^2$. The equation becomes $u^2 - 7u + 12 = 0$, or $(u-3)(u-4)=0$. Since $u=3$ or $u=4$, we have $x^2=3$ (so $x=\pm\sqrt{3}$) and $x^2=4$ (so $x=\pm2$).
• Solve $(x-1)^2 + 5(x-1) + 6 = 0$. Let $u=x-1$. The equation becomes $u^2+5u+6=0$, or $(u+2)(u+3)=0$. Since $u=-2$ or $u=-3$, we have $x-1=-2$ (so $x=-1$) and $x-1=-3$ (so $x=-2$).
• Solve $x^{\frac{2}{3}} - 2x^{\frac{1}{3}} - 8 = 0$. Let $u=x^{\frac{1}{3}}$. The equation becomes $u^2 - 2u - 8 = 0$, or $(u-4)(u+2)=0$. Since $u=4$ or $u=-2$, we have $x^{\frac{1}{3}}=4$ (so $x=64$) and $x^{\frac{1}{3}}=-2$ (so $x=-8$).

Explanation

Some equations don't look like quadratics, but they behave like them. If one exponent is double the other (like $x^4$ and $x^2$), you can use a substitution to turn it into a familiar $au^2+bu+c=0$ form. Solve for $u$, then substitute back.

Property

When solving rational equations, the process of clearing denominators by multiplying by the Least Common Denominator (LCD) can lead to a quadratic equation.
Once the rational equation is transformed into a quadratic, it can be solved by factoring or the quadratic formula.
It is crucial to check all potential solutions in the original equation, as some may be extraneous solutions that make a denominator zero.

Examples

• Solve $\frac{x}{x-3} + \frac{1}{x+1} = \frac{12}{x^2-2x-3}$. Multiply by the LCD $(x-3)(x+1)$ to get $x(x+1) + (x-3) = 12$. This simplifies to $x^2+2x-15=0$, or $(x+5)(x-3)=0$. The solution is $x=-5$, as $x=3$ is extraneous.
• Solve $\frac{6}{x} + x = 5$. Multiply by the LCD, $x$, to get $6 + x^2 = 5x$. Rearrange into $x^2-5x+6=0$, which factors to $(x-2)(x-3)=0$. The solutions are $x=2$ and $x=3$, both of which are valid.
• Solve $\frac{2x^2}{x^2-4} = \frac{x}{x-2} - \frac{3}{x+2}$. Multiply by the LCD $(x-2)(x+2)$ to get $2x^2 = x(x+2) - 3(x-2)$. This simplifies to $x^2+x-6=0$, or $(x+3)(x-2)=0$. The only valid solution is $x=-3$, as $x=2$ is extraneous.

Explanation

Sometimes, when you clear the fractions in a rational equation, you're left with a quadratic equation. Solve this new quadratic as you normally would, but be extra careful to check your answers. Any solution that makes an original denominator zero is invalid.