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?

Issa inti L-ebda kalkulatur ma jista' jsolvi din il-problema, iżda l-biċċiet tagħha jistgħu jiġu kkalkulati. Ipprova waħda minn hawn taħt, jew ittajpja tiegħek stess.

⌨ Tip ✎ Iġbed dan

Problema matematika tiegħek

π Ħoll

Prattika (40)

Ipprova kull waħda fuq il-karta l-ewwel.Iżvela t-tweġiba biex tiċċekkja; dawk ivverifikati jistgħu jinfetħu fis-solver għal kull pass.

1. Factor by substitution: \({y}^{4}-{y}^{2}-20.\)

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

   \(\left({y}^{2}+4\right)\left({y}^{2}-5\right)\)

2. Factor by substitution: \({(y-4)}^{2}+8(y-4)+15.\)

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

   \((y-1)(y+1)\)

3. Simplify: ⓐ \({x}^{\frac{1}{2}}\cdot {x}^{\frac{1}{4}}\) ⓑ \({({x}^{\frac{1}{3}})}^{2}\) ⓒ \({({x}^{-1})}^{2}.\)

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

   ⓐ \({x}^{\frac{3}{4}}\); ⓑ \({x}^{\frac{2}{3}}\); ⓒ \({x}^{-2}\)

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

   Xogħol permezz tagħha Ħjiel

5. Solve: \({x}^{4}-6{x}^{2}+8=0\).

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

   \(x=\sqrt{2},x=\text{-}\sqrt{2},x=2,x=-2\)

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

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

   \(x=\sqrt{7},x=\text{-}\sqrt{7},x=2,x=-2\)

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

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

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

8. Solve: \({(x-5)}^{2}+6(x-5)+8=0.\)

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

   \(x=3,x=1\)

9. Solve: \({(y-4)}^{2}+8(y-4)+15=0.\)

   Xogħol permezz tagħha Ħjiel

   Jiżvelaw it-tweġiba

   \(y=-1,y=1\)

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    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:

11. Solve: \(x-7\sqrt{x}+12=0.\)

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    \(x=9,x=16\)

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    \(x=4,x=16\)

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    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:

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    \(x=-8,x=343\)

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    \(x=81,x=625\)

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    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:

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    \(x=\frac{2}{5},x=\frac{3}{4}\)

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

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

    Xogħol permezz tagħha Ħjiel

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

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

    Xogħol permezz tagħha Ħjiel

23. \(2{x}^{4}-5{x}^{2}+3=0\)

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

24. \(4{x}^{4}-5{x}^{2}+1=0\)

    Xogħol permezz tagħha Ħjiel

25. \(2{x}^{4}-7{x}^{2}+3=0\)

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

26. \(3{x}^{4}-14{x}^{2}+8=0\)

    Xogħol permezz tagħha Ħjiel

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

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

    Xogħol permezz tagħha Ħjiel

29. \({(3y+2)}^{2}+(3y+2)-6=0\)

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

30. \({(5y-1)}^{2}+3(5y-1)-28=0\)

    Xogħol permezz tagħha Ħjiel

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

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

    Xogħol permezz tagħha Ħjiel

33. \(2{({x}^{2}-5)}^{2}-5({x}^{2}-5)+2=0\)

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

34. \(2{({x}^{2}-5)}^{2}-7({x}^{2}-5)+6=0\)

    Xogħol permezz tagħha Ħjiel

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    \(x=25\)

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

    Xogħol permezz tagħha Ħjiel

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

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

    \(x=4\)

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

    Xogħol permezz tagħha Ħjiel

39. \(6x+\sqrt{x}-2=0\)

    Xogħol permezz tagħha Ħjiel

    Jiżvelaw it-tweġiba

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

40. \(6x+\sqrt{x}-1=0\)

    Xogħol permezz tagħha Ħjiel

Żomm tiegħek stess li jaħdmu

Kont b'xejn iżid noti fuq kull lezzjoni, rekord ta' dak li lest, problemi solvuti tiegħek f'post wieħed, u għalliem inti tista' tistaqsi dwar din il-paġna.Il-matematika nnifisha hija miftuħa għal kulħadd, log fil jew le.

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Simboli użati hawnhekk

Taptap kwalunkwe simbolu għad-definizzjoni sħiħa, stampa, u dak li kull ittra fiha tfisser.

\sqrt{x}square root, n-th root
The non-negative number whose square (n-th power) is x.iimaginary unit
i² = −1.\pmplus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.\neqnot equal
The two sides are different.\leqless/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 xlogarithm, natural log
The exponent b must be raised to for x; ln uses base e.\mathbb{N}number sets
Naturals, integers, rationals, reals, complex numbers.

Kif għandek tuża: Solve Equations in Quadratic Form

1. Solve equations in quadratic form
2. Identify a substitution that will put the equation in quadratic form.
3. Rewrite the equation with the substitution to put it in quadratic form.
4. Solve the quadratic equation for
5. Substitute the original variable back into the results, using the substitution.
6. Solve for the original variable.
7. Check the solutions.
8. How to solve equations in quadratic form.

Mistoqsijiet li n-nies jistaqsu

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.

Partijiet minn din il-paġna huma adattati minn OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Kondensat u mill-ġdid spjegat hawn; żbalji huma tagħna. Minħabba li s-sors tiegħu huwa liċenzjat għal użu mhux kummerċjali, din il-lezzjoni hija b'xejn, ma tbigħ xejn, u hija maqsuma taħt l-istess liċenzja.

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