Solve Equations in Quadratic Form

Solve Equations in Quadratic Form

Sometimes when we factored trinomials, the trinomial did not appear to be in the $ax^2+bx+c$ form. So we factored by substitution allowing us to make it fit the $ax^2+bx+c$ form. We used the standard $u$ for the substitution.

To factor the expression $x^4-4x^2-5$, we noticed the variable part of the middle term is $x^2$ and its square, $x^4$, is the variable part of the first term. (We know $(x^2)^2=x^4$.) So we let $u=x^2$ and factored.

Similarly, sometimes an equation is not in the $ax^2+bx+c=0$ form but looks much like a quadratic equation. Then, we can often make a thoughtful substitution that will allow us to make it fit the $ax^2+bx+c=0$ form. If we can make it fit the form, we can then use all of our methods to solve quadratic equations.

Notice that in the quadratic equation $ax^2+bx+c=0$, the middle term has a variable, $x$, and its square, $x^2$, is the variable part of the first term. Look for this relationship as you try to find a substitution.

Again, we will use the standard $u$ to make a substitution that will put the equation in quadratic form. If the substitution gives us an equation of the form $ax^2+bx+c=0$, we say the original equation was in quadratic form.

The next example shows the steps for solving an equation in quadratic form.

Example. Solve $6x^4-7x^2+2=0$.

Solution.

Since $(x^2)^2=x^4$, we let $u=x^2$.

There are four solutions:

Check all four solutions. We will show one check here.

We leave the other checks to you.

We summarize the steps to solve an equation in quadratic form.

In the next example, 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 $ax^2+bx+c$ form.

Example. Solve $(x-2)^2+7(x-2)+12=0$.

Solution.

Check:

In the next example, we notice that $(\sqrt{x})^2=x$. Also, remember that when we square both sides of an equation, we may introduce extraneous roots. Be sure to check your answers!

Example. Solve $x-3\sqrt{x}+2=0$.

Solution.

The $\sqrt{x}$ in the middle term is squared in the first term, $(\sqrt{x})^2=x$. If we let $u=\sqrt{x}$ and substitute, our trinomial will be in $ax^2+bx+c=0$ form.

Check:

Substitutions for rational exponents can also help us solve an equation in quadratic form. Think of the properties of exponents as you begin the next example.

Example. Solve $x^{\tfrac{2}{3}}-2x^{\tfrac{1}{3}}-24=0$.

Solution.

The $x^{\tfrac{1}{3}}$ in the middle term is squared in the first term, $(x^{\tfrac{1}{3}})^2=x^{\tfrac{2}{3}}$. If we let $u=x^{\tfrac{1}{3}}$ and substitute, our trinomial will be in $ax^2+bx+c=0$ form.

Check:

In the next example, we need to keep in mind the definition of a negative exponent as well as the properties of exponents.

Example. Solve $3x^{-2}-7x^{-1}+2=0$.

Solution.

The $x^{-1}$ in the middle term is squared in the first term, $(x^{-1})^2=x^{-2}$. If we let $u=x^{-1}$ and substitute, our trinomial will be in $ax^2+bx+c=0$ form.

Check:

Key terms. An equation is in quadratic form when a substitution can rewrite it in the form $au^2+bu+c=0$, where $a\ne0$.