Objective: Solve equations that are quadratic in form by substitution to create a quadratic equation.
We have seen three different ways to solve quadratics: factoring, completing the square, and the quadratic formula. A quadratic is any equation of the form \(0=ax^2+bx+c\text{,}\) however, we can use the skills learned to solve quadratics to solve problems with higher (or sometimes lower) powers if the equation is in what is called quadratic form.
An equation is in quadratic form if one of the exponents on a variable is double the exponent on the same variable somewhere else in the equation. If this is the case we can create a new variable, set it equal to the variable with smallest exponent. When we substitute this into the equation we will have a quadratic equation we can solve.
\begin{align*}
x^4-13x^2+36=0 \amp \quad \text{Quadratic form, one exponent, $4$, double the other, $2$} \\
y=x^2 \amp\quad \text{New variable equal to the variable with smaller exponent} \\
y^2=x^4 \amp\quad \text{Square both sides} \\
y^2-13y+36=0 \amp\quad \text{Substitute} y \text{for} x^2 \text{ and} y^2 \text{for} x^4 \\
(y-9)(y-4)=0 \amp\quad \text{Solve. We can solve this equation by factoring} \\
y-9=0 \text{ or } y-4=0 \amp\quad \text{Set each factor equal to zero } \\
\underline{+9 +9} \qquad\underline{+4 +4} \amp\quad \text{Solutions for $y$ , need $x$. We will use} y = x^2 \text{equation} \\
y=9 \text{ or } y=4 \amp\quad \text{Solutions for $y$,
need $x$. We will use } y=x^2 \text{equation} \\
9=x^2 \text{ or } 4=x^2 \amp\quad \text{Substitute values for $y$} \\
\pm \sqrt{9}=\sqrt{x^2} \text{ or } \pm \sqrt{4}=\sqrt{x^2} \amp\quad \text{Solve using the even root property, simplify roots} \\
x= \pm 3, \pm 2 \amp\quad \text{Our Solutions}
\end{align*}
\begin{align*}
a^{-2}-a^{-1}-6=0 \amp \quad \text{Quadratic form, one exponent, $- 2$, is double the other, $- 1$} \\
b=a^{-1} \amp\quad \text{Make a new variable equal to the variable with lowest exponent} \\
b^2=a^{-2} \amp\quad \text{Square both sides} \\
b^2-b-6=0 \amp \quad \text{Substitute } b^2 \text{ for } a^{-2} \text{ and } b \text{ for } a^{-1} \\
(b-3)(b+2)=0 \amp\quad \text{Solve. We will solve by factoring} \\
b-3=0 \text{ or } b+2=0 \amp\quad \text{Set each factor equal to $0$} \\
\underline{+3 +3} \qquad \underline{-2 -2} \amp \quad \text{Solve each equation} \\
b=3 \text{ or } b=-2 \amp\quad \text{Solutions for
$b$, still need $a$, substitute into } b = a^{-1} \\
3=a^{-1} \text{ or } -2=a^{-1} \amp\quad \text{Raise both sides to $- 1$ power} \\
3^{-1}=a \text{ or } (-2)^{-1} \amp \quad \text{Simplify negative exponents} \\
a= \frac{1}{3}, - \frac{1}{2} \amp\quad \text{Our Solution}
\end{align*}
Just as with regular quadratics, these problems will not always have rational solutions. We also can have irrational or complex solutions to our equations.
When we create a new variable for our substitution, it won’t always be equal to just another variable. We can make our substitution variable equal to an expression as shown in the next example.
