# 1.6: Equations Quadratic in Form

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## Solving Equations in Quadratic Form

Equations in quadratic form are equations with three terms.  The term with the highest exponent (called the leading term) has a power that is a multiple of $$2$$, the middle term has an exponent that is one-half the exponent of the leading term, and the third term is a constant. An equation that is quadratic in form can be written in the form $$a u^{2}+b u+c=0$$ where $$u$$ represents an algebraic expression.

A few examples of these equations include:

\begin{aligned} x^4−5x^2+4 &=0 \stackrel{u=x^2} {\color{Cerulean}{\Longrightarrow}} u^2-5u+4 = 0 \\ x^6+7x^3−8 &=0 \stackrel{u=x^3} {\color{Cerulean}{\Longrightarrow}} u^2+7u-8=0 \\ x^{\tfrac{2}{3}} +4x^{\tfrac{1}{3}}+2 &=0 \stackrel{x^{\tfrac{1}{3}}} {\color{Cerulean}{\Longrightarrow}} u^2+4u+2=0 \\ \left(\frac{t+2}{t}\right)^{2}+8\left(\frac{t+2}{t}\right)+7&=0 \stackrel{u=\frac{t+2}{t}}{\color{Cerulean}{\Longrightarrow}} \color{black}{u^{2}}+8 u+7=0 \\ x-3 \sqrt{x}-10&=0 \stackrel{u=\sqrt{x}} {\color{Cerulean}{\Longrightarrow}} \color{black}{ u^{2}}-3 u-10=0 \\ 3 y^{-2}+7 y^{-1}-6&=0 \stackrel{u=y^{-1}} {\color{Cerulean}{\Longrightarrow}} \color{black}{ 3 u^{2}}+7 u-6=0 \end{aligned}

In each example, doubling the exponent of the middle term equals the exponent on the leading term.
If an equation can be expressed in quadratic form, then it can be solved by any of the techniques used to solve ordinary quadratic equations. For example, consider the following fourth-degree polynomial equation,

$$x^{4}-4 x^{2}-32=0$$

If we let $$u = x^{2}$$ then $$u^{2} = \left(x^{2}\right)^{2}=x^{4}$$ and we can write

$$\begin{array}{r}{x^{4}-4 x^{2}-32=0 \color{Cerulean}{\Rightarrow}\left(\color{Cerulean}{x^{2}}\right)^{\color{black}{2}}-4\left(\color{Cerulean}{x^{2}}\right)\color{black}{-}32=0} \\ \color{Cerulean}{\downarrow\quad\quad\:\:\: \downarrow\quad\quad\quad\quad\:\:} \\ {u^{2}\quad-\:\:4 u\:\:-32=0}\end{array}$$

This substitution transforms the equation into a familiar quadratic equation in terms of $$u$$ which, in this case, can be solved by factoring.

\begin{aligned} u^{2}-4 u-32 &=0 \\(u-8)(u+4) &=0 \\ u=8 \quad \text { or } \quad u &=-4 \end{aligned}

Since $$u=x^{2}$$ we can back substitute and then solve for $$x$$.

\begin{aligned} &u=8 \quad \text{or} &u=-4 \\ & \color{Cerulean}{\downarrow}&\color{Cerulean}{ \downarrow} \\ &x^{2}=8 & x^{2}=-4\\ &x=\pm \sqrt{8} &x=\pm\sqrt{-4} \\ &x=\pm 2 \sqrt{2} &x=\pm2i\end{aligned}

Therefore, the equation $$x^{4}-4 x^{2}-32=0$$ has four solutions $$\{\pm 2 \sqrt{2}, \pm 2 i\}$$, two real and two complex. This technique is called a u-substitution, and can often be used to solve equations that are quadratic in form. Howto: Solve an equation that is quadratic in form.

1. Confirm the equation is quadratic in form and rewrite it.
1. The equation is quadratic in form if the exponent on the leading term is double the exponent on the middle term.
2. Substitute  $$u$$ for the variable portion of the middle term and rewrite the equation in the form  $$a u^{2}+b u+c=0$$ .
2. Solve using one of the usual methods for solving a quadratic.
3. Back substitute. Replace $$u$$ with the original term.
4. Solve the resulting equations for the original variable. Check for domain restrictions or extraneous solutions if necessary.

Example $$\PageIndex{1}$$: Solving a Fourth-degree Equation in Quadratic Form

Solve this fourth-degree equation: $$3x^4−2x^2−1=0$$.

Solution

Step 1. This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let $$u =x^2$$. Rewrite the equation in $$u$$.

$3u^2−2u−1=0 \nonumber$

Step 2. Now solve the quadratic.

\begin{align*} 3u^2-2u-1&= 0\\ (3u+1)(u-1)&= 0 \end{align*}
$$\begin{array} {r|r} 3u+1= 0 & u-1= 0 \\ 3u= -1 & u= 1 \\ u= -\dfrac{1}{3} & \\ \end{array}$$

Step 3. Replace $$u$$ with the original term.

$$x^2= -\dfrac{1}{3} \qquad x^2= 1$$

Step 4. Solve for the original variable

$$\begin{array} {r|r} x^2= -\dfrac{1}{3} & x^2= 1 \\ x= \pm i\sqrt{\dfrac{1}{3}} & x= \pm 1 \\ \end{array}$$

Neither the original equation nor the solution process pose any restrictions to $$x$$ so the solution set is:  $$\Large\{$$  $$±i\sqrt{\tfrac{1}{3}}, ±1 \Large\}$$

Example $$\PageIndex{2}$$: Solving an Equation in Quadratic Form Containing a Binomial

Solve the equation in quadratic form: $${(x+2)}^2+11(x+2)−12=0$$.

Solution

This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting $$u =x+2$$. Then rewrite the equation in $$u$$.

$$u^2+11u-12= 0$$

Solve for $$u$$ using the zero-factor property

$$(u+12)(u-1)= 0$$
$$\begin{array} {r|r} u+12=0 & u-1=0 \\ u=-12 & u=1 \\ \end{array}$$

Replace $$u$$ with the original expression and solve for $$x$$

$$\begin{array} {r|r} x+2=-12 & x+2=1 \\ x=-14 & x=-1 \\ \end{array}$$

Neither the original equation nor the solution process pose any restrictions to $$x$$ so the solution set is:  $$\{−14, −1 \}$$.

Example $$\PageIndex{3}$$:Solving an Equation in Quadratic Form Containing a Radical

Solve: $$x-2 \sqrt{x}-8=0$$.

Solution

This is a radical equation that can be written in quadratic form. If we let $$u=\sqrt{x}$$ then $$u^{2}=(\sqrt{x})^{2}=x$$ and we can write

\begin{aligned}x-2 \sqrt{x}-8&=0\\ \color{Cerulean}{\downarrow\:\:\quad \downarrow\quad\:\:\:\:} \\ u^{2}\:\:-2 u-8&=0\end{aligned}

Solve for $$u$$.

$$u^{2}-2 u-8=0$$
$$(u-4)(u+2)=0$$
$$u=4$$ or $$u=-2$$

Back substitute $$u=\sqrt{x}$$ and solve for $$x$$.

$$\begin{array}{c}{\sqrt{x}=4 \quad \text { or } \quad \sqrt{x}=-2} \\ {(\sqrt{x})^{2}=(4)^{2} \quad(\sqrt{x})^{2}=(-2)^{2}} \\ {x=16} \quad\quad\quad\quad {x=4}\end{array}$$

The original equation poses the restriction that $$x \ge 0$$ which neither solution obtained violates, but recall that squaring both sides of an equation introduces the possibility of extraneous solutions. Therefore we must check our potential solutions.

$$\begin{array}{r | r} {\color{OliveGreen}{Check}} \:x = 16 & {\color{OliveGreen}{Check}}\:x = 4\\ {\sqrt{x} = 4} & {\sqrt{x} = -2} \\ \sqrt{ { \color{OliveGreen}{16}} } = 4 & \sqrt{ {\color{OliveGreen}{4}} } = -2 \\ {4 = 4 \color{Cerulean}{✓}} & {\quad\quad\quad\:\: 2 = -2 \color{red}{✗}} \end{array}$$

Because $$x=4$$ is extraneous, there is only one solution, $$x=16$$.
Solution Set:  $$\{16 \}$$.

Example $$\PageIndex{4}$$: Solving an Equation in Quadratic Form Containing Rational Exponents

Solve: $$x^{2 / 3}-3 x^{1 / 3}-10=0$$.

Solution

If we let $$u=x^{1 / 3}$$ then $$u^{2}=\left(x^{1 / 3}\right)^{2}=x^{2 / 3}$$ and we can write

$$\begin{array}{c}{x^{2 / 3}-3 x^{1 / 3}-10=0} \\ \color{Cerulean}{\downarrow \quad\quad\downarrow\:\:\quad\quad\quad\quad\:} \\ {u^{2}\:\:\:-\:3 u-10=0}\end{array}$$

Solve for $$u$$.

$$u^{2}-3 u-10=0$$
$$(u-5)(u+2)=0$$
$$u=5 \quad$$ or $$\quad u=-2$$

Back substitute $$u=x^{1 / 3}$$ and solve for $$x$$.

\begin{aligned} x^{1 / 3} &=5 \quad \text { or } \quad x^{1 / 3}=-2 \\\left(x^{1 / 3}\right)^{3} &=(5)^{3} \quad\left(x^{1 / 3}\right)^{3}=(-2)^{3} \\ x &=125 \quad\quad\quad\:\:\: x=-8 \end{aligned}

The domain of the original equation does not have any restrictions. Furthermore, the solution process involved cubing, not squaring. Squaring can introduce extraneous solutions, but cubing does not, so a check is not necessary.

Solution Set:  $$\{-8, 125 \}$$.

Example $$\PageIndex{5}$$: Solving an Equation in Quadratic Form Containing Negative Exponents

Solve: $$3 y^{-2}+7 y^{-1}-6=0$$

Solution

If we let $$u=y^{-1}$$, then $$u^{2}=\left(y^{-1}\right)^{2}=y^{-2}$$ and we can write

$$\begin{array}{c}{3 y^{-2}+7 y^{-1}-6=0} \\ \color{Cerulean}{\downarrow \quad\:\:\:\:\:\downarrow\quad\quad\:\:\:\;\:} \\ {3 u^{2}+7 u-6=0}\end{array}$$

Solve for $$u$$.

$$3 u^{2}+7 u-6=0$$
$$(3 u-2)(u+3)=0$$
$$u=\dfrac{2}{3} \quad$$ or $$\quad u=-3$$

Back substitute $$u=y^{-1}$$ and solve for $$y$$.

\begin{aligned} y^{-1} &=\frac{2}{3} \text { or } y^{-1}=-3 \\ \frac{1}{y} &=\frac{2}{3} \quad \quad\frac{1}{y}=-3 \\ y &=\frac{3}{2} \quad\quad y=-\frac{1}{3} \end{aligned}

The original equation is a rational equation where $$y ≠ 0$$. In this case, the solutions are not restrictions.

The solution set is  $$\large\{ -\tfrac{1}{3}, \tfrac{3}{2} \large\}$$.

Example $$\PageIndex{6}$$: Solving an Equation in Quadratic Form Containing Rational Expressions

Solve: $$\left(\dfrac{t+2}{t}\right)^{2}+8\left(\dfrac{t+2}{t}\right)+7=0$$

Solution

If we let $$u=\dfrac{t+2}{t}$$, then $$u^{2}=\left(\dfrac{t+2}{t}\right)^{2}$$ and we can write

$$\begin{array}{c}{\left(\dfrac{t+2}{t}\right)^{2}+8\left(\dfrac{t+2}{t}\right)+7=0}\\ \color{Cerulean}{\downarrow \quad\quad\:\:\;\:\downarrow\quad\quad\quad\quad} \\ {u^{2}+\quad 8 u \quad+7=0}\end{array}$$

Solve for $$u$$.

$$u^{2}+8 u+7=0$$
$$(u+1)(u+7)=0$$
$$u=-1 \quad$$ or $$\quad u=-7$$

Back substitute $$u=\dfrac{t+2}{t}$$, and solve for $$t$$.

$$\begin{array}{rlcrl} \dfrac{t+2}{t} &= -1 & \text{ or } & \dfrac{t+2}{t} &= -7 \\ t+2 &= -t & & t+2 &= -7t \\ 2t &= -2 & & 8t &= -2 \\ t &= -1 & & t &= -\dfrac{1}{4} \\ \end{array}$$

The domain restriction to the equation is $$t \ne 0$$, so both solutions work and the solution set is  $${\large\{} -1, -\tfrac{1}{4} {\large\}}$$.

Example $$\PageIndex{7}$$: Solving an Equation in Quadratic Form Containing a Radical Rational Expression

Solve: $$\sqrt{\dfrac {x+10}{x-6} } +\sqrt{\dfrac {x-6}{x+10} } -\dfrac{34}{15}=0$$

Solution

If we let $$u=\sqrt{\dfrac {x+10}{x-6} }$$, then $$\dfrac{1}{u} = \sqrt{\dfrac {x-6}{x+10}}$$ and we can write

$$\begin{array}{c}{\sqrt{\dfrac {x+10}{x-6} } +\sqrt{\dfrac {x-6}{x+10} } -\dfrac{34}{15}=0} \\ \color{Cerulean}{\downarrow \quad\:\:\:\:\:\downarrow\quad\quad\:\:\:\;\:} \\ \quad\:{u+\dfrac{1}{u}-\dfrac{34}{15}=0}\end{array}$$

Solve for $$u$$. First multiply by 15$$u$$

$$15u^2+15-34u=0$$
$$15u^2-34u+15=0$$
$$(3u-5)(5u-3)=0$$
$$u=\frac{5}{3} \quad$$ or $$\quad u=\frac{3}{5}$$

Back substitute $$u=\sqrt{\dfrac {x+10}{x-6} }$$ and solve for $$x$$.

\begin{align*} \sqrt{\frac {x+10}{x-6} } &=\frac{5}{3} & \text { or } & & \sqrt{\frac {x+10}{x-6} } &=\frac{3}{5} \\[4pt] \frac {x+10}{x-6} &=\frac{25}{9} & & & \dfrac {x+10}{x-6} &=\frac{9}{25} \\[4pt] 9x + 90 &=25x-150 & & & 25x+250&=9x-54 \\[4pt] 240 &=16x & & & 16x&=-304 \\[4pt] x &=15 & & & x &=-19 \end{align*}

Check. The restrictions to the domain of the original equation are $$x > 6$$ and $$x < -10$$ and do not apply to the solutions found, but the solution process involved squaring, so the answers must be checked.

Using $$x = 15$$,  we obtain $$\sqrt{\dfrac{x+10}{x-6} } = \sqrt{\dfrac {15+10}{15-6} } = \sqrt{\dfrac {25}{9} } = \dfrac{5}{3} \quad {\color{Cerulean}{✓}}$$
Using $$x = -19$$,  we obtain $$\sqrt{\dfrac{x+10}{x-6} } = \sqrt{\dfrac {-19+10}{-19-6}} = \sqrt{\dfrac {-9}{-25}} = \dfrac{3}{5} \quad {\color{Cerulean}{✓}}$$

The solution set is $$\{ 15, -19 \}$$.

So far all of the examples were of equations that factor. As we know, not all quadratic equations factor. If this is the case, we use the quadratic formula.

Example $$\PageIndex{8}$$: Solving a Fourth-degree Equation in Unfactorable Quadratic Form

Solve $$x^{4}-10 x^{2}+23=0$$.

Solution

If we let $$u=x^{2}$$, then $$u^{2}=\left(x^{2}\right)^{2}=x^{4}$$ and we can write

$$\begin{array}{l}{x^{4}-10 x^{2}+23=0}\\ \color{Cerulean}{\downarrow \quad\:\:\: \downarrow} \\ {u^{2}-10 u+23=0}\end{array}$$

This equation does not factor; therefore, use the quadratic formula to find the solutions for $$u$$. Here $$a = 1, b = −10$$, and $$c = 23$$.

\begin{aligned} u &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(-10) \pm \sqrt{(-10)^{2}-4(1)(23)}}{2(1)} \\ &=\frac{10 \pm \sqrt{8}}{2} \\ &=\frac{10 \pm 2 \sqrt{2}}{2} \\ &=5 \pm \sqrt{2} \end{aligned}

Therefore, $$u=5\pm\sqrt{2}$$. Now back substitute $$u=x^{2}$$ and solve for $$x$$.

$$\begin{array}{c}{u=5-\sqrt{2} \quad \text { or } \quad u=5+\sqrt{2}}\\\color{Cerulean}{\downarrow\quad\quad\quad\quad\quad\quad\quad\downarrow\quad\quad\quad\quad} \\ {x^{2}=5-\sqrt{2} \quad\quad\:\:\: x^{2}=5+\sqrt{2}} \\ {x=\pm \sqrt{5-\sqrt{2}} \quad\quad\: x=\pm \sqrt{5+\sqrt{2}}}\end{array}$$

There are no domain restrictions nor potential extraneous solutions so the solution set is $${\large\{} \pm \sqrt{5+\sqrt{2}},\pm \sqrt{5-\sqrt{2}} {\large\}}$$

If approximate values to the nearest hundredth are desired, the four approximate solutions are

$$x=-\sqrt{5-\sqrt{2}} \approx-1.89 \quad x=-\sqrt{5+\sqrt{2}} \approx-2.53$$
$$x=\sqrt{5-\sqrt{2}} \approx 1.89 \quad x=\sqrt{5+\sqrt{2}} \approx 2.53$$ Try It $$\PageIndex{6}$$

Solve using substitution:

 a.   $$x^4−8x^2−9=0$$ b.   $${(x−5)}^2−4(x−5)−21=0$$ c.   $$12 x^{-2}-16 x^{-1}+5=0$$
 a.   $$x=−3,3,−i,i$$ b.   $$x=2,x=12$$ c.    $$\{\frac{6}{5}, 2 \}$$