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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?

À vous Aucune calculatrice ne tranche celle-ci, mais ses éléments sont calculables. Essayez-en un ci-dessous, ou tapez le vôtre.

Exercices (40)

Essayez chacun d'abord sur papier. Révélez la réponse pour vérifier; les vérifiés peuvent être ouverts dans le solveur pour chaque étape.

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

    Révèle la réponse

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

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

    Révèle la réponse

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

    Révèle la réponse

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

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

    Révèle la réponse
    Prepare for the substitution.
    Let \(u=x-2\) and substitute.
    Solve by factoring.
    Replace \(u\) with \(x-2.\)
    Solve for \(x.\)
    Check:
  8. Solve: \({(x-5)}^{2}+6(x-5)+8=0.\)

    Révèle la réponse

    \(x=3,x=1\)

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

    Révèle la réponse

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

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

    Révèle la réponse

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

    Révèle la réponse

    \(x=9,x=16\)

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

    Révèle la réponse

    \(x=4,x=16\)

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

    Révèle la réponse

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

    Révèle la réponse

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

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

    Révèle la réponse

    \(x=81,x=625\)

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

    Révèle la réponse

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

    Révèle la réponse

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

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

    Révèle la réponse

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

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

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

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

    Révèle la réponse

    \(x=25\)

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

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

    Révèle la réponse

    \(x=4\)

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

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

    Révèle la réponse

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

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

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Comment faire: 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.

Questions posées par les gens

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.

Certaines parties de cette page sont adaptées de OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensés et réexpliqués ici ; les erreurs sont les nôtres. Comme sa source est autorisée à des fins non commerciales, cette leçon est gratuite, ne vend rien et est partagée avec la même licence.

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