2.27: Solving Equations in Quadratic Form

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Solving 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 $x^{4}-4 x^{2}-5$, 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 $\left(x^{2}\right)^{2}=x^{4}$.) So we let $u=x^{2}$ and factored.

	$x^{4}-4 x^{2}-5$
	$\left(\color{red}x^2 \color{black} \right)^{2}-4\left( \color{red}x^{2} \color{black}\right)-5$
Let $u=x^{2}$ and substitute.	$\color{red}u \color{black}^{2}-4 \color{red}u \color{black}-5$
Factor the trinomial.	$(u+1)(u-5)$
Replace $u$ with $x^{2}$.	$\left( \color{red}x^{2} \color{black} + 1\right)\left( \color{red}x^2 \color{black}-5\right)$

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 of quadratic form.

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

Example $\PageIndex{1}$

Solve: $6 x^{4}-7 x^{2}+2=0$

Solution:

Step 1: Identify a substitution that will put the equation in quadratic form.	Since $\left(x^{2}\right)^{2}=x^{4}$, we let $u=x^{2}$.	$6 x^{4}-7 x^{2}+2=0$
Step 2: Rewrite the equation with the substitution to put it in quadratic form.	Rewrite to prepare for the substitution. Substitute $u=x^{2}$.	$\begin{aligned}6\color{black}{\left(\color{red}{x^{2}}\right)}^{2}-7\color{red}{ x^{2}}\color{black}{+}2&=0 \\ \color{black}{6 \color{red}{u}^{2}}-7 \color{red}{u}\color{black}{+}2&=0\end{aligned}$
Step 3: Solve the quadratic equation for $u$.	We can solve by factoring. Use the Zero Product Property.	$\begin{aligned}(2 u-1)(3 u-2) &=0 \\ 2 u-1=0, 3 u-2&=0 \\ 2 u =1,3 u&=2 \\ u =\frac{1}{2} u&=\frac{2}{3} \end{aligned}$
Step 4: Substitute the original variable back into the results, using the substitution.	Replace $u$ with $x^{2}$.	$x^{2}=\frac{1}{2} \quad x^{2}=\frac{2}{3}$
Step 5: Solve for the original variable.	Solve for $x$, using the Square Root Property.	$\begin{array}{ll}{x=\pm \sqrt{\frac{1}{2}}} & {x=\pm \sqrt{\frac{2}{3}}} \\ {x=\pm \frac{\sqrt{2}}{2}} & {x=\pm \frac{\sqrt{6}}{3}}\end{array}$ There are four solutions. $\begin{array}{ll}{x=\frac{\sqrt{2}}{2}} & {x=\frac{\sqrt{6}}{3}} \\ {x=-\frac{\sqrt{2}}{2}} & {x=-\frac{\sqrt{6}}{3}}\end{array}$
Step 6: Check the solutions.	Check all four solutions. We will show one check here.	$\begin{aligned}x&=\frac{\sqrt{2}}{2} \\ 6 x^{4}-7 x^{2}+2&=0 \\ 6\left(\frac{\sqrt{2}}{2}\right)^{4}-7\left(\frac{\sqrt{2}}{2}\right)^{2}+2 &\stackrel{?}{=} 0\\ 6\left(\frac{4}{16} \right)-7\left(\frac{2}{4} \right)^{2}+2&\stackrel{?}{=}0 \\ \frac{3}{2}-\frac{7}{2}+\frac{4}{2}&\stackrel{?}{=}0 \\ 0&=0 \end{aligned}$ We leave the other checks to you!

You Try $\PageIndex{1}$

Solve: $m^{4}-6 m^{2}+8=0$

Answer: $m=\sqrt{2}, m=-\sqrt{2}, m=2, m=-2$

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

Solve Equations in Quadratic Form

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

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 $a x^{2}+b x+c$ form.

Example $\PageIndex{2}$

Solve: $(x-2)^{2}+7(x-2)+12=0$

Solution:

	$(x-2)^2+7(x-2)+12=0$ $.$
Prepare for the substitution.	${\color{Red} (x-2)}^2+7{\color{Red} (x-2)}+12=0$
Let $u=x-2$ and substitute.	${\color{Red} u}^2+7 {\color{Red} u}+12=0$
Solve by factoring.	$(u+3)(u+4)=0$ $u+3=0, \quad u+4=0$ $u=-3, \quad u=-4$
Replace $u$ with $x-2$.	$x-2=-3, x-2=-4$
Solve for $x$.	\| $x=-1, \quad x=-2$
Check:	$\begin{array}{l} x=-1 \\ (x-2)^2+7(x-2)+12=0 \\ ({\color{Red} -1}-2)^2+7({\color{Red} -1}-2)+12 \stackrel{?}{=} 0 \\ (-3)^2+7(-3)+12 \stackrel{?}{=} 0 \\ 9-21+12 \stackrel{?}{=} 0 \\ 0=0\end{array}$ $\begin{aligned} x & =-2 \\ (x-2)^2+7(x-2)+12 & =0 \\ ({\color{Red} -2}-2)^2+7({\color{Red} -2}-2)+12 & \stackrel{?}{=} 0 \\ (-4)^2+7(-4)+12 & \stackrel{?}{=} 0 \\ 16-28+12 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}$

You Try $\PageIndex{2}$

Solve: $(a-5)^{2}+6(a-5)+8=0$

Answer: $a=3, a=1$

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 $\PageIndex{3}$

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 $a x^{2}+b x+c=0$ form.

	$x-3 \sqrt{x}+2=0$
Rewrite the trinomial to prepare for the substitution.	$({\color{Pink} \sqrt{x}})^2-3 {\color{Pink} \sqrt{x}}+2=0$
Let $u=\sqrt{x}$ and substitute.	${\color{Pink} u}^2-3 {\color{Pink} u}+2=0$
Solve by factoring.	$(u-2)(u-1)=0$ $u-2=0, \quad u-1=0$ $u=2, \quad u=1$
Replace $u$ with $\sqrt{x}$.	$\sqrt{x}=2, \quad \sqrt{x}=1$
Solve for $x$, by squaring both sides.	$x=4, \quad x=1$
Check:	$\begin{aligned} x & =-2 \\ (x-2)^2+7(x-2)+12 & =0 \\ ({\color{Red} -2}-2)^2+7({\color{Red} -2}-2)+12 & \stackrel{?}{=} 0 \\ (-4)^2+7(-4)+12 & \stackrel{?}{=} 0 \\ 16-28+12 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}$ $\begin{aligned} x & =1 \\ x-3 \sqrt{x}+2 & =0 \\ {\color{Red} 1}-3 \sqrt{ 1}+2 & \stackrel{?}{=} 0 \\ 1-3+2 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}$

You Try $\PageIndex{3}$

Solve: $p-6 \sqrt{p}+8=0$

Answer: $p=4, p=16$

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 $\PageIndex{4}$

Solve: $x^{\frac{2}{3}}-2 x^{\frac{1}{3}}-24=0$

Solution:

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

	$x^{\frac{2}{3}}-2 x^{\frac{1}{3}}-24=0$
Rewrite the trinomial to prepare for the substitution.	$\left ( {\color{Red} x^{\frac{1}{3}}} \right )^{2}-2 {\color{Red} x^{\frac{1}{3}}}-24=0$
Let $u=x^{\frac{1}{3}}$	${\color{Red} u}^2-2 {\color{Red} u}-24=0$
Solve by factoring.	$(u-6)(u+4)=0$ $u-6=0, \quad u+4=0$ $u=6, \quad u=-4$
Replace $u$ with $x^{\frac{1}{3}}$.	$x^{\frac{1}{3}}=6, \quad x^{\frac{1}{3}}=-4$
Solve for $x$ by cubing both sides.	$\left(x^{\frac{1}{3}}\right)^{3}=(6)^{3}, \quad\left(x^{\frac{1}{3}}\right)^{3}=(-4)^{3}$ $x=216, \quad x=-64$
Check:	$\begin{array}{l} x=216 \\ x^{\frac{2}{3}}-2 x^{\frac{1}{3}}-24=0 \\ ({\color{Red} 216})^{\frac{2}{3}}-2({\color{Red} 216})^{\frac{1}{3}}-24 \stackrel{?}{=} 0 \\ 36-12-24 \stackrel{?}{=} 0 \\ 0=0 \checkmark\end{array}$ $\begin{array}{l} x=-64 \\ x^{\frac{2}{3}}-2 x^{\frac{1}{3}}-24=0 \\ ({\color{Red} -64})^{\frac{2}{3}}-2({\color{Red} -64})^{\frac{1}{3}}-24 \stackrel{?}{=} 0 \\ 16+8-24 \stackrel{?}{=} 0 \\ 0=0 \checkmark\end{array}$

You Try $\PageIndex{4}$

Solve: $x^{\frac{1}{2}}+8 x^{\frac{1}{4}}+15=0$

Answer: $x=81, x=625$

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

Example $\PageIndex{5}$

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

Solution:

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

	$3 x^{-2}-7 x^{-1}+2=0$
Rewrite the trinomial to prepare for the substitution.	$3\left({\color{Red} x^{-1}}\right)^2-7\left({\color{Red} x^{-1}}\right)+2=0$
Let $u=x^{-1}$ and substitute.	$3{\color{Red} u}^2-7 {\color{Red} u}+2=0$
Solve by factoring.	$(3 u-1)(u-2)=0$
	$3 u-1=0, \quad u-2=0$
	$u=\frac{1}{3}, \quad u=2$
Replace $u$ with $x^{-1}$.	$x^{-1}=\dfrac{1}{3}, \quad x^{-1}=2$
Solve for $x$ by taking the reciprocal since $x^{-1}=\frac{1}{x}$.	$x=3, \quad x=\frac{1}{2}$
Check:	$\begin{aligned} x & =3 \\ 3 x^{-2}-7 x^{-1}+2 & =0 \\ 3({\color{Pink} 3})^{-2}-7({\color{Pink} 3})^{-1}+2 & \stackrel{?}{=} 0 \\ 3\left(\frac{1}{9}\right)-7\left(\frac{1}{3}\right)+2 & \stackrel{?}{=} 0 \\ \frac{1}{3}-\left(\frac{7}{3}\right)+\frac{6}{3} & \stackrel{?}{=} 0 \\ 0 & =0 .\end{aligned}$ $\begin{aligned} x & =\frac{1}{2} \\ 3 x^{-2}-7 x^{-1}+2 & =0 \\ 3\left({\color{Pink} \frac{1}{2}} \right)^{-2}-7 \left({\color{Pink} \frac{1}{2}}\right)^{-1}+2 & \stackrel{?}{=} 0 \\ 3(4)-7(2)+2 & \stackrel{?}{=} 0 \\ 12-14+2 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}$

You Try $\PageIndex{5}$

Solve: $6 v^{-2}-23 v^{-1}+20=0$

Answer: $v=\frac{2}{5}, v=\frac{3}{4}$

	\(x^{4}-4 x^{2}-5\)
	\(\left(\color{red}x^2 \color{black} \right)^{2}-4\left( \color{red}x^{2} \color{black}\right)-5\)
Let \(u=x^{2}\) and substitute.	\(\color{red}u \color{black}^{2}-4 \color{red}u \color{black}-5\)
Factor the trinomial.	\((u+1)(u-5)\)
Replace \(u\) with \(x^{2}\).	\(\left( \color{red}x^{2} \color{black} + 1\right)\left( \color{red}x^2 \color{black}-5\right)\)

Step 1: Identify a substitution that will put the equation in quadratic form.	Since \(\left(x^{2}\right)^{2}=x^{4}\), we let \(u=x^{2}\).	\(6 x^{4}-7 x^{2}+2=0\)
Step 2: Rewrite the equation with the substitution to put it in quadratic form.	Rewrite to prepare for the substitution. Substitute \(u=x^{2}\).	\(\begin{aligned}6\color{black}{\left(\color{red}{x^{2}}\right)}^{2}-7\color{red}{ x^{2}}\color{black}{+}2&=0 \\ \color{black}{6 \color{red}{u}^{2}}-7 \color{red}{u}\color{black}{+}2&=0\end{aligned}\)
Step 3: Solve the quadratic equation for \(u\).	We can solve by factoring. Use the Zero Product Property.	\(\begin{aligned}(2 u-1)(3 u-2) &=0 \\ 2 u-1=0, 3 u-2&=0 \\ 2 u =1,3 u&=2 \\ u =\frac{1}{2} u&=\frac{2}{3} \end{aligned}\)
Step 4: Substitute the original variable back into the results, using the substitution.	Replace \(u\) with \(x^{2}\).	\(x^{2}=\frac{1}{2} \quad x^{2}=\frac{2}{3}\)
Step 5: Solve for the original variable.	Solve for \(x\), using the Square Root Property.	\(\begin{array}{ll}{x=\pm \sqrt{\frac{1}{2}}} & {x=\pm \sqrt{\frac{2}{3}}} \\ {x=\pm \frac{\sqrt{2}}{2}} & {x=\pm \frac{\sqrt{6}}{3}}\end{array}\) There are four solutions. \(\begin{array}{ll}{x=\frac{\sqrt{2}}{2}} & {x=\frac{\sqrt{6}}{3}} \\ {x=-\frac{\sqrt{2}}{2}} & {x=-\frac{\sqrt{6}}{3}}\end{array}\)
Step 6: Check the solutions.	Check all four solutions. We will show one check here.	\(\begin{aligned}x&=\frac{\sqrt{2}}{2} \\ 6 x^{4}-7 x^{2}+2&=0 \\ 6\left(\frac{\sqrt{2}}{2}\right)^{4}-7\left(\frac{\sqrt{2}}{2}\right)^{2}+2 &\stackrel{?}{=} 0\\ 6\left(\frac{4}{16} \right)-7\left(\frac{2}{4} \right)^{2}+2&\stackrel{?}{=}0 \\ \frac{3}{2}-\frac{7}{2}+\frac{4}{2}&\stackrel{?}{=}0 \\ 0&=0 \end{aligned}\) We leave the other checks to you!

	\((x-2)^2+7(x-2)+12=0\) $.$
Prepare for the substitution.	\({\color{Red} (x-2)}^2+7{\color{Red} (x-2)}+12=0\)
Let \(u=x-2\) and substitute.	\({\color{Red} u}^2+7 {\color{Red} u}+12=0\)
Solve by factoring.	\((u+3)(u+4)=0\) \(u+3=0, \quad u+4=0\) \(u=-3, \quad u=-4\)
Replace \(u\) with \(x-2\).	\(x-2=-3, x-2=-4\)
Solve for \(x\).	\| \(x=-1, \quad x=-2\)
Check:	\(\begin{array}{l} x=-1 \\ (x-2)^2+7(x-2)+12=0 \\ ({\color{Red} -1}-2)^2+7({\color{Red} -1}-2)+12 \stackrel{?}{=} 0 \\ (-3)^2+7(-3)+12 \stackrel{?}{=} 0 \\ 9-21+12 \stackrel{?}{=} 0 \\ 0=0\end{array}\) \(\begin{aligned} x & =-2 \\ (x-2)^2+7(x-2)+12 & =0 \\ ({\color{Red} -2}-2)^2+7({\color{Red} -2}-2)+12 & \stackrel{?}{=} 0 \\ (-4)^2+7(-4)+12 & \stackrel{?}{=} 0 \\ 16-28+12 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}\)

	\(x-3 \sqrt{x}+2=0\)
Rewrite the trinomial to prepare for the substitution.	\(({\color{Pink} \sqrt{x}})^2-3 {\color{Pink} \sqrt{x}}+2=0\)
Let \(u=\sqrt{x}\) and substitute.	\({\color{Pink} u}^2-3 {\color{Pink} u}+2=0\)
Solve by factoring.	\((u-2)(u-1)=0\) \(u-2=0, \quad u-1=0\) \(u=2, \quad u=1\)
Replace \(u\) with \(\sqrt{x}\).	\(\sqrt{x}=2, \quad \sqrt{x}=1\)
Solve for \(x\), by squaring both sides.	\(x=4, \quad x=1\)
Check:	\(\begin{aligned} x & =-2 \\ (x-2)^2+7(x-2)+12 & =0 \\ ({\color{Red} -2}-2)^2+7({\color{Red} -2}-2)+12 & \stackrel{?}{=} 0 \\ (-4)^2+7(-4)+12 & \stackrel{?}{=} 0 \\ 16-28+12 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}\) \(\begin{aligned} x & =1 \\ x-3 \sqrt{x}+2 & =0 \\ {\color{Red} 1}-3 \sqrt{ 1}+2 & \stackrel{?}{=} 0 \\ 1-3+2 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}\)

	\(x^{\frac{2}{3}}-2 x^{\frac{1}{3}}-24=0\)
Rewrite the trinomial to prepare for the substitution.	\(\left ( {\color{Red} x^{\frac{1}{3}}} \right )^{2}-2 {\color{Red} x^{\frac{1}{3}}}-24=0\)
Let \(u=x^{\frac{1}{3}}\)	\({\color{Red} u}^2-2 {\color{Red} u}-24=0\)
Solve by factoring.	\((u-6)(u+4)=0\) \(u-6=0, \quad u+4=0\) \(u=6, \quad u=-4\)
Replace \(u\) with \(x^{\frac{1}{3}}\).	\(x^{\frac{1}{3}}=6, \quad x^{\frac{1}{3}}=-4\)
Solve for \(x\) by cubing both sides.	\(\left(x^{\frac{1}{3}}\right)^{3}=(6)^{3}, \quad\left(x^{\frac{1}{3}}\right)^{3}=(-4)^{3}\) \(x=216, \quad x=-64\)
Check:	\(\begin{array}{l} x=216 \\ x^{\frac{2}{3}}-2 x^{\frac{1}{3}}-24=0 \\ ({\color{Red} 216})^{\frac{2}{3}}-2({\color{Red} 216})^{\frac{1}{3}}-24 \stackrel{?}{=} 0 \\ 36-12-24 \stackrel{?}{=} 0 \\ 0=0 \checkmark\end{array}\) \(\begin{array}{l} x=-64 \\ x^{\frac{2}{3}}-2 x^{\frac{1}{3}}-24=0 \\ ({\color{Red} -64})^{\frac{2}{3}}-2({\color{Red} -64})^{\frac{1}{3}}-24 \stackrel{?}{=} 0 \\ 16+8-24 \stackrel{?}{=} 0 \\ 0=0 \checkmark\end{array}\)

Search

Example \(\PageIndex{1}\)

You Try \(\PageIndex{1}\)

Solve Equations in Quadratic Form

Example \(\PageIndex{2}\)

You Try \(\PageIndex{2}\)

Example \(\PageIndex{3}\)

You Try \(\PageIndex{3}\)

Example \(\PageIndex{4}\)

You Try \(\PageIndex{4}\)

Example \(\PageIndex{5}\)

You Try \(\PageIndex{5}\)

	\(3 x^{-2}-7 x^{-1}+2=0\)
Rewrite the trinomial to prepare for the substitution.	\(3\left({\color{Red} x^{-1}}\right)^2-7\left({\color{Red} x^{-1}}\right)+2=0\)
Let \(u=x^{-1}\) and substitute.	\(3{\color{Red} u}^2-7 {\color{Red} u}+2=0\)
Solve by factoring.	\((3 u-1)(u-2)=0\)
	\(3 u-1=0, \quad u-2=0\)
	\(u=\frac{1}{3}, \quad u=2\)
Replace \(u\) with \(x^{-1}\).	\(x^{-1}=\dfrac{1}{3}, \quad x^{-1}=2\)
Solve for \(x\) by taking the reciprocal since \(x^{-1}=\frac{1}{x}\).	\(x=3, \quad x=\frac{1}{2}\)
Check:	\(\begin{aligned} x & =3 \\ 3 x^{-2}-7 x^{-1}+2 & =0 \\ 3({\color{Pink} 3})^{-2}-7({\color{Pink} 3})^{-1}+2 & \stackrel{?}{=} 0 \\ 3\left(\frac{1}{9}\right)-7\left(\frac{1}{3}\right)+2 & \stackrel{?}{=} 0 \\ \frac{1}{3}-\left(\frac{7}{3}\right)+\frac{6}{3} & \stackrel{?}{=} 0 \\ 0 & =0 .\end{aligned}\) \(\begin{aligned} x & =\frac{1}{2} \\ 3 x^{-2}-7 x^{-1}+2 & =0 \\ 3\left({\color{Pink} \frac{1}{2}} \right)^{-2}-7 \left({\color{Pink} \frac{1}{2}}\right)^{-1}+2 & \stackrel{?}{=} 0 \\ 3(4)-7(2)+2 & \stackrel{?}{=} 0 \\ 12-14+2 & \stackrel{?}{=} 0 \\ 0 & =0\end{aligned}\)