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Solve Equations in Quadratic Form

Solve equations in quadratic form

Solve Equations in Quadratic Form

Sometimes when we factored trinomials, the trinomial did not appear to be in the ax² + bx + c form. So we factored by substitution allowing us to make it fit the ax² + bx + c form. We used the standard \(u\) for the substitution.

To factor the expression x⁴ − 4x² − 5, we noticed the variable part of the middle term is x² and its square, x⁴, is the variable part of the first term. (We know \({({x}^{2})}^{2}={x}^{4}.\)) So we let u = x² and factored.



Let \(u={x}^{2}\) and substitute.
Factor the trinomial.
Replace u with \({x}^{2}\).

Similarly, sometimes an equation is not in the ax² + 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 ax² + 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 ax² + bx + c = 0, the middle term has a variable, x, and its square, x², 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 ax² + 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.

How to Solve Equations in Quadratic Form

Try it.

Solve: \(6{x}^{4}-7{x}^{2}+2=0\)

Solution

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

Example

Try it.

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

Solution


Prepare for the substitution.
Let \(u=x-2\) and substitute.
Solve by factoring.
Replace \(u\) with \(x-2.\)
Solve for \(x.\)
Check:

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

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.

Solve Equations in Quadratic Form

Solve Equations in Quadratic Form

In the following exercises, solve.

Try it.

\({x}^{4}-7{x}^{2}+12=0\)

Solution

\(x=\pm \sqrt{3},x=\pm 2\)

Try it.

\({x}^{4}-9{x}^{2}+18=0\)

Try it.

\({x}^{4}-13{x}^{2}-30=0\)

Solution

\(x=\pm \sqrt{15},x=\pm \sqrt{2}i\)

Try it.

\({x}^{4}+5{x}^{2}-36=0\)

Try it.

\(2{x}^{4}-5{x}^{2}+3=0\)

Solution

\(x=\pm 1,x=\frac{\pm \sqrt{6}}{2}\)

Try it.

\(4{x}^{4}-5{x}^{2}+1=0\)

Try it.

\(2{x}^{4}-7{x}^{2}+3=0\)

Solution

\(x=\pm \sqrt{3},x=\pm \frac{\sqrt{2}}{2}\)

Try it.

\(3{x}^{4}-14{x}^{2}+8=0\)

Try it.

\({(x-3)}^{2}-5(x-3)-36=0\)

Solution

\(x=-1,x=12\)

Try it.

\({(x+2)}^{2}-3(x+2)-54=0\)

Try it.

\({(3y+2)}^{2}+(3y+2)-6=0\)

Solution

\(x=-\frac{5}{3},x=0\)

Try it.

\({(5y-1)}^{2}+3(5y-1)-28=0\)

Try it.

\({({x}^{2}+1)}^{2}-5({x}^{2}+1)+4=0\)

Solution

\(x=0,x=\pm \sqrt{3}\)

Try it.

\({({x}^{2}-4)}^{2}-4({x}^{2}-4)+3=0\)

Try it.

\(2{({x}^{2}-5)}^{2}-5({x}^{2}-5)+2=0\)

Solution

\(x=\pm \frac{\sqrt{22}}{2},x=\pm \sqrt{7}\)

Try it.

\(2{({x}^{2}-5)}^{2}-7({x}^{2}-5)+6=0\)

Try it.

\(x-\sqrt{x}-20=0\)

Solution

\(x=25\)

Try it.

\(x-8\sqrt{x}+15=0\)

Try it.

\(x+6\sqrt{x}-16=0\)

Solution

\(x=4\)

Try it.

\(x+4\sqrt{x}-21=0\)

Try it.

\(6x+\sqrt{x}-2=0\)

Solution

\(x=\frac{1}{4}\)

Try it.

\(6x+\sqrt{x}-1=0\)

Try it.

\(10x-17\sqrt{x}+3=0\)

Solution

\(x=\frac{1}{25},x=\frac{9}{4}\)

Try it.

\(12x+5\sqrt{x}-3=0\)

Try it.

\({x}^{\frac{2}{3}}+9{x}^{\frac{1}{3}}+8=0\)

Solution

\(x=-1,x=-512\)

Try it.

\({x}^{\frac{2}{3}}-3{x}^{\frac{1}{3}}=28\)

Try it.

\({x}^{\frac{2}{3}}+4{x}^{\frac{1}{3}}=12\)

Solution

\(x=8,x=-216\)

Try it.

\({x}^{\frac{2}{3}}-11{x}^{\frac{1}{3}}+30=0\)

Try it.

\(6{x}^{\frac{2}{3}}-{x}^{\frac{1}{3}}=12\)

Solution

\(x=\frac{27}{8},x=-\frac{64}{27}\)

Try it.

\(3{x}^{\frac{2}{3}}-10{x}^{\frac{1}{3}}=8\)

Try it.

\(8{x}^{\frac{2}{3}}-43{x}^{\frac{1}{3}}+15=0\)

Solution

\(x=\frac{27}{512},x=125\)

Try it.

\(20{x}^{\frac{2}{3}}-23{x}^{\frac{1}{3}}+6=0\)

Try it.

\(x-8{x}^{\frac{1}{2}}+7=0\)

Solution

\(x=1,x=49\)

Try it.

\(2x-7{x}^{\frac{1}{2}}=15\)

Try it.

\(6{x}^{-2}+13{x}^{-1}+5=0\)

Solution

\(x=-2,x=-\frac{3}{5}\)

Try it.

\(15{x}^{-2}-26{x}^{-1}+8=0\)

Try it.

\(8{x}^{-2}-2{x}^{-1}-3=0\)

Solution

\(x=-2,x=\frac{4}{3}\)

Try it.

\(15{x}^{-2}-4{x}^{-1}-4=0\)

Try it.

Explain how to recognize an equation in quadratic form.

Solution

Answers will vary.

Try it.

Explain the procedure for solving an equation in quadratic form.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Factor by substitution: \({y}^{4}-{y}^{2}-20.\)
If you missed this problem, review .

Bonisa impendulo

\(\left({y}^{2}+4\right)\left({y}^{2}-5\right)\)
Factor by substitution: \({(y-4)}^{2}+8(y-4)+15.\)
If you missed this problem, review .

Bonisa impendulo

\((y-1)(y+1)\)
Simplify: ⓐ \({x}^{\frac{1}{2}}\cdot {x}^{\frac{1}{4}}\) ⓑ \({({x}^{\frac{1}{3}})}^{2}\) ⓒ \({({x}^{-1})}^{2}.\)
If you missed this problem, review .

Bonisa impendulo

ⓐ \({x}^{\frac{3}{4}}\); ⓑ \({x}^{\frac{2}{3}}\); ⓒ \({x}^{-2}\)
Solve: \(6{x}^{4}-7{x}^{2}+2=0\)
Solve: \({x}^{4}-6{x}^{2}+8=0\).

Bonisa impendulo

\(x=\sqrt{2},x=\text{-}\sqrt{2},x=2,x=-2\)
Solve: \({x}^{4}-11{x}^{2}+28=0\).

Bonisa impendulo

\(x=\sqrt{7},x=\text{-}\sqrt{7},x=2,x=-2\)
Solve: \({(x-2)}^{2}+7(x-2)+12=0.\)

Bonisa impendulo

Prepare for the substitution.
Let \(u=x-2\) and substitute.
Solve by factoring.

Replace \(u\) with \(x-2.\)
Solve for \(x.\)
Check:
Solve: \({(x-5)}^{2}+6(x-5)+8=0.\)

Bonisa impendulo

\(x=3,x=1\)
Solve: \({(y-4)}^{2}+8(y-4)+15=0.\)

Bonisa impendulo

\(y=-1,y=1\)
Solve: \(x-3\sqrt{x}+2=0.\)

Bonisa impendulo

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 ax² + bx + c = 0 form.

Rewrite the trinomial to prepare for the substitution.
Let \(u=\sqrt{x}\) and substitute.
Solve by factoring.

Replace u with \(\sqrt{x}.\)
Solve for x, by squaring both sides.
Check:
Solve: \(x-7\sqrt{x}+12=0.\)

Bonisa impendulo

\(x=9,x=16\)
Solve: \(x-6\sqrt{x}+8=0.\)

Bonisa impendulo

\(x=4,x=16\)

Solve: \({x}^{\frac{2}{3}}-2{x}^{\frac{1}{3}}-24=0.\)

Bonisa impendulo

The \({x}^{\frac{1}{3}}\) in the middle term is squared in the first term \({({x}^{\frac{1}{3}})}^{2}={x}^{\frac{2}{3}}.\) If we let \(u={x}^{\frac{1}{3}}\) and substitute, our trinomial will be in ax² + bx + c = 0 form.


Rewrite the trinomial to prepare for the substitution.
Let \(u={x}^{\frac{1}{3}}\) and substitute.
Solve by factoring.
Replace u with \({x}^{\frac{1}{3}}.\)
Solve for \(x\) by cubing both sides.
Check:

Solve: \({x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}-14=0.\)

Bonisa impendulo

\(x=-8,x=343\)
Solve: \({x}^{\frac{1}{2}}-8{x}^{\frac{1}{4}}+15=0.\)

Bonisa impendulo

\(x=81,x=625\)

Solve: \(3{x}^{-2}-7{x}^{-1}+2=0.\)

Bonisa impendulo

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 ax² + bx + c = 0 form.


Rewrite the trinomial to prepare for the substitution.
Let \(u={x}^{-1}\) and substitute.
Solve by factoring.


Replace u with \({x}^{-1}.\)
Solve for \(x\) by taking the reciprocal since \({x}^{-1}=\frac{1}{x}.\)
Check:

Solve: \(8{x}^{-2}-10{x}^{-1}+3=0.\)

Bonisa impendulo

\(x=\frac{4}{3}x=2\)
Solve: \(6{x}^{-2}-23{x}^{-1}+20=0.\)

Bonisa impendulo

\(x=\frac{2}{5},x=\frac{3}{4}\)
\({x}^{4}-7{x}^{2}+12=0\)

Bonisa impendulo

\(x=\pm \sqrt{3},x=\pm 2\)
\({x}^{4}-9{x}^{2}+18=0\)
\({x}^{4}-13{x}^{2}-30=0\)

Bonisa impendulo

\(x=\pm \sqrt{15},x=\pm \sqrt{2}i\)
\({x}^{4}+5{x}^{2}-36=0\)
\(2{x}^{4}-5{x}^{2}+3=0\)

Bonisa impendulo

\(x=\pm 1,x=\frac{\pm \sqrt{6}}{2}\)
\(4{x}^{4}-5{x}^{2}+1=0\)
\(2{x}^{4}-7{x}^{2}+3=0\)

Bonisa impendulo

\(x=\pm \sqrt{3},x=\pm \frac{\sqrt{2}}{2}\)
\(3{x}^{4}-14{x}^{2}+8=0\)
\({(x-3)}^{2}-5(x-3)-36=0\)

Bonisa impendulo

\(x=-1,x=12\)
\({(x+2)}^{2}-3(x+2)-54=0\)
\({(3y+2)}^{2}+(3y+2)-6=0\)

Bonisa impendulo

\(x=-\frac{5}{3},x=0\)
\({(5y-1)}^{2}+3(5y-1)-28=0\)
\({({x}^{2}+1)}^{2}-5({x}^{2}+1)+4=0\)

Bonisa impendulo

\(x=0,x=\pm \sqrt{3}\)
\({({x}^{2}-4)}^{2}-4({x}^{2}-4)+3=0\)
\(2{({x}^{2}-5)}^{2}-5({x}^{2}-5)+2=0\)

Bonisa impendulo

\(x=\pm \frac{\sqrt{22}}{2},x=\pm \sqrt{7}\)
\(2{({x}^{2}-5)}^{2}-7({x}^{2}-5)+6=0\)
\(x-\sqrt{x}-20=0\)

Bonisa impendulo

\(x=25\)
\(x-8\sqrt{x}+15=0\)
\(x+6\sqrt{x}-16=0\)

Bonisa impendulo

\(x=4\)
\(x+4\sqrt{x}-21=0\)
\(6x+\sqrt{x}-2=0\)

Bonisa impendulo

\(x=\frac{1}{4}\)
\(6x+\sqrt{x}-1=0\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

imaginary unit
i² = −1.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Equations 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
Substitute the original variable back into the results, using the substitution.
Solve for the original variable.
Check the solutions.
How to solve equations in quadratic form.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Zama ngokwakho

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Rewrite the trinomial to prepare for the substitution.
Let \(u=\sqrt{x}\) and substitute.
Solve by factoring.
Replace u with \(\sqrt{x}.\)
Solve for x, by squaring both sides.
Check: