Lesson 9.4: Solve Equations in Quadratic Form

New Concept

In this lesson, you'll master solving equations in "quadratic form." These equations can be transformed into the familiar $au^2 + bu + c = 0$ format using a clever substitution, allowing you to apply your existing quadratic-solving skills.

What’s next

Soon, you'll tackle interactive examples and practice problems to master this powerful substitution technique across different types of equations.

Property

An equation is of quadratic form if a substitution gives us an equation of the form $au^2+bu+c=0$. We look for a relationship where the variable part of the first term is the square of the variable part of the middle term. For example, in $6x^4 - 7x^2 + 2 = 0$, we see $(x^2)^2 = x^4$.

How to solve equations in quadratic form:

1. Identify a substitution that will put the equation in quadratic form.
2. Rewrite the equation with the substitution to put it in quadratic form.
3. Solve the quadratic equation for $u$.
4. Substitute the original variable back into the results, using the substitution.
5. Solve for the original variable.
6. Check the solutions.

Examples

• Solve $x^4 - 10x^2 + 9 = 0$. Let $u=x^2$. This gives $u^2 - 10u + 9 = 0$, so $(u-9)(u-1)=0$. Thus $u=9$ or $u=1$. Substituting back, $x^2=9$ gives $x=\pm3$, and $x^2=1$ gives $x=\pm1$.

Property

In an equation like $(x-2)^2 + 7(x-2) + 12 = 0$, the binomial in the middle term, $(x-2)$, is squared in the first term. If we let $u = x-2$ and substitute, our trinomial will be in $au^2 + bu + c$ form.

Examples

• Solve $(x+1)^2 + 5(x+1) + 6 = 0$. Let $u=x+1$. This gives $u^2+5u+6=0$, so $(u+2)(u+3)=0$. Thus $u=-2$ or $u=-3$. Substituting back, $x+1=-2$ gives $x=-3$, and $x+1=-3$ gives $x=-4$.

• Solve $(2x-1)^2 - (2x-1) - 12 = 0$. Let $u=2x-1$. This gives $u^2-u-12=0$, so $(u-4)(u+3)=0$. Thus $u=4$ or $u=-3$. Substituting back, $2x-1=4$ gives $x=\frac{5}{2}$, and $2x-1=-3$ gives $x=-1$.

Property

In an equation like $x - 3\sqrt{x} + 2 = 0$, the $\sqrt{x}$ in the middle term is squared in the first term, since $(\sqrt{x})^2 = x$. If we let $u = \sqrt{x}$ and substitute, our trinomial will be in $au^2+bu+c=0$ form. When we square both sides of an equation, we may introduce extraneous roots, so be sure to check your answers.

Examples

• Solve $x - 5\sqrt{x} + 4 = 0$. Let $u=\sqrt{x}$. This gives $u^2-5u+4=0$, so $(u-1)(u-4)=0$. Thus $u=1$ or $u=4$. Substituting back, $\sqrt{x}=1$ gives $x=1$, and $\sqrt{x}=4$ gives $x=16$. Both solutions are valid.

• Solve $x + 2\sqrt{x} - 15 = 0$. Let $u=\sqrt{x}$. This gives $u^2+2u-15=0$, so $(u+5)(u-3)=0$. Thus $u=-5$ or $u=3$. Since $\sqrt{x}$ cannot be negative, $u=-5$ is extraneous. For $u=3$, $\sqrt{x}=3$ gives $x=9$.

Property

In an equation like $x^{\frac{2}{3}} - 2x^{\frac{1}{3}} - 24 = 0$, the $x^{\frac{1}{3}}$ in the middle term is squared in the first term, since $(x^{\frac{1}{3}})^2 = x^{\frac{2}{3}}$. If we let $u = x^{\frac{1}{3}}$ and substitute, our trinomial will be in $au^2+bu+c$ form. To solve for $x$, raise both sides to the reciprocal power.

Examples

• Solve $x^{\frac{2}{3}} + 2x^{\frac{1}{3}} - 8 = 0$. Let $u=x^{\frac{1}{3}}$. This gives $u^2+2u-8=0$, so $(u+4)(u-2)=0$. Thus $u=-4$ or $u=2$. Substituting back, $x^{\frac{1}{3}}=-4$ gives $x=(-4)^3=-64$, and $x^{\frac{1}{3}}=2$ gives $x=2^3=8$.

• Solve $x^{\frac{1}{2}} - 5x^{\frac{1}{4}} + 6 = 0$. Let $u=x^{\frac{1}{4}}$. This gives $u^2-5u+6=0$, so $(u-2)(u-3)=0$. Thus $u=2$ or $u=3$. Substituting back, $x^{\frac{1}{4}}=2$ gives $x=2^4=16$, and $x^{\frac{1}{4}}=3$ gives $x=3^4=81$.

Property

In an equation like $3x^{-2} - 7x^{-1} + 2 = 0$, the $x^{-1}$ in the middle term is squared in the first term, since $(x^{-1})^2 = x^{-2}$. If we let $u = x^{-1}$ and substitute, our trinomial will be in $au^2+bu+c=0$ form. Solve for $x$ by taking the reciprocal, since $x^{-1} = \frac{1}{x}$.

Examples

• Solve $x^{-2} - x^{-1} - 6 = 0$. Let $u=x^{-1}$. This gives $u^2-u-6=0$, so $(u-3)(u+2)=0$. Thus $u=3$ or $u=-2$. Substituting back, $x^{-1}=3$ gives $x=\frac{1}{3}$, and $x^{-1}=-2$ gives $x=-\frac{1}{2}$.

• Solve $2x^{-2} + 5x^{-1} - 3 = 0$. Let $u=x^{-1}$. This gives $2u^2+5u-3=0$, so $(2u-1)(u+3)=0$. Thus $u=\frac{1}{2}$ or $u=-3$. Substituting back, $x^{-1}=\frac{1}{2}$ gives $x=2$, and $x^{-1}=-3$ gives $x=-\frac{1}{3}$.