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?

Ugbu a ị Enweghị kalkulata na-ewepụ nke a, ma ọ bụ na ọ nwere ike ijikwa ya. Jiri otu n'okpuru, mọọbụ pịa nke gị.

⌨ Ụdị ✎ Debata ya

Nsogbu igodo gị

π Wepụ

Nhazi (40)

Nwapụta n'ime akwụkwọ n'oge mbụ. Gosi nzaghachi ka ịlelee ya; ndị ahụ e gosipụtara nwere ike ịbụ ndị a mepere n'ime onye nrụpụta maka nzọụkwụ ọbụla.

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

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

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

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

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

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

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

   Ọrụ site na ya Ndụ

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

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

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

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

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

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

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

   \(x=3,x=1\)

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

   Ọrụ site na ya Ndụ

   Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

    \(x=9,x=16\)

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

    \(x=4,x=16\)

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

    \(x=81,x=625\)

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

    \(x=25\)

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

    \(x=4\)

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

    Ọrụ site na ya Ndụ

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

    Ọrụ site na ya Ndụ

    Gosi nzaghachi

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

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

    Ọrụ site na ya Ndụ

Kwụsị ọrụ gị

Akwụkwọ ndebanye n'efu na-egbakwunye ntọala n'ụdị ọmụmụ ihe ọbụla, ndepụta nke ihe ị ga-eme, nsogbu ị ga-ewepụ n'otu ebe, na onye nkụzi ị ga-ajụ banyere ihuakwụkwọ a. Akwụkwọ ndepụta ihe bụ nke ọbụla mepere maka onye ọbụla, ọbụla ma ọ bụ na ọ bụghị.

Akaụntụ Nbanye

Aịkọn ndị ahụ ejirila ebe a

Tinye ọbụla simbụl maka nkọwa zuru ezu, inyogo, na ihe ọbụla ọbụla na-egosi.

\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.

Olee otú: 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.

Ajụjụ ndị mmadụ jụrụ

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.

Oge nke ihuakwụkwọ a bụ nke a gbanwere site na OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). E kọwaara ya na ebe a; nsogbu bụ nke anyị. N'ihi na ọbjektị ya bụ nke e ji akwụ ụgwọ maka ojiji na-enweghị ọrụ, agụụ a bụ nke ọbụla, ọ na-ereghị ihe ọ bụla, ma e mewe ya n'okpuru ikike ahụ.

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