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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.
- 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.
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?
Sekarang Anda Tidak ada kalkulator yang bisa menuntaskan soal ini, tetapi bagian-bagiannya dapat dihitung. Coba salah satu di bawah ini, atau ketik soal Anda sendiri.
Latihan (40)
Coba kerjakan setiap soal di kertas terlebih dahulu. Buka jawabannya untuk memeriksa; soal yang terverifikasi dapat dibuka di pemecah soal untuk melihat setiap langkah.
-
Factor by substitution: \({y}^{4}-{y}^{2}-20.\)
Tampilkan jawaban
\(\left({y}^{2}+4\right)\left({y}^{2}-5\right)\)
-
Factor by substitution: \({(y-4)}^{2}+8(y-4)+15.\)
Tampilkan jawaban
\((y-1)(y+1)\)
-
Simplify: ⓐ \({x}^{\frac{1}{2}}\cdot {x}^{\frac{1}{4}}\) ⓑ \({({x}^{\frac{1}{3}})}^{2}\) ⓒ \({({x}^{-1})}^{2}.\)
Tampilkan jawaban
ⓐ \({x}^{\frac{3}{4}}\); ⓑ \({x}^{\frac{2}{3}}\); ⓒ \({x}^{-2}\)
-
Solve: \(6{x}^{4}-7{x}^{2}+2=0\)
-
Solve: \({x}^{4}-6{x}^{2}+8=0\).
Tampilkan jawaban
\(x=\sqrt{2},x=\text{-}\sqrt{2},x=2,x=-2\)
-
Solve: \({x}^{4}-11{x}^{2}+28=0\).
Tampilkan jawaban
\(x=\sqrt{7},x=\text{-}\sqrt{7},x=2,x=-2\)
-
Solve: \({(x-2)}^{2}+7(x-2)+12=0.\)
Tampilkan jawaban
Prepare for the substitution. Let \(u=x-2\) and substitute. Solve by factoring. Replace \(u\) with \(x-2.\) Solve for \(x.\) Check: -
Solve: \({(x-5)}^{2}+6(x-5)+8=0.\)
Tampilkan jawaban
\(x=3,x=1\)
-
Solve: \({(y-4)}^{2}+8(y-4)+15=0.\)
Tampilkan jawaban
\(y=-1,y=1\)
-
Solve: \(x-3\sqrt{x}+2=0.\)
Tampilkan jawaban
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: \(x-7\sqrt{x}+12=0.\)
Tampilkan jawaban
\(x=9,x=16\)
-
Solve: \(x-6\sqrt{x}+8=0.\)
Tampilkan jawaban
\(x=4,x=16\)
-
Solve: \({x}^{\frac{2}{3}}-2{x}^{\frac{1}{3}}-24=0.\)
Tampilkan jawaban
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: \({x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}-14=0.\)
Tampilkan jawaban
\(x=-8,x=343\)
-
Solve: \({x}^{\frac{1}{2}}-8{x}^{\frac{1}{4}}+15=0.\)
Tampilkan jawaban
\(x=81,x=625\)
-
Solve: \(3{x}^{-2}-7{x}^{-1}+2=0.\)
Tampilkan jawaban
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: \(8{x}^{-2}-10{x}^{-1}+3=0.\)
Tampilkan jawaban
\(x=\frac{4}{3}x=2\)
-
Solve: \(6{x}^{-2}-23{x}^{-1}+20=0.\)
Tampilkan jawaban
\(x=\frac{2}{5},x=\frac{3}{4}\)
-
\({x}^{4}-7{x}^{2}+12=0\)
Tampilkan jawaban
\(x=\pm \sqrt{3},x=\pm 2\)
-
\({x}^{4}-9{x}^{2}+18=0\)
-
\({x}^{4}-13{x}^{2}-30=0\)
Tampilkan jawaban
\(x=\pm \sqrt{15},x=\pm \sqrt{2}i\)
-
\({x}^{4}+5{x}^{2}-36=0\)
-
\(2{x}^{4}-5{x}^{2}+3=0\)
Tampilkan jawaban
\(x=\pm 1,x=\frac{\pm \sqrt{6}}{2}\)
-
\(4{x}^{4}-5{x}^{2}+1=0\)
-
\(2{x}^{4}-7{x}^{2}+3=0\)
Tampilkan jawaban
\(x=\pm \sqrt{3},x=\pm \frac{\sqrt{2}}{2}\)
-
\(3{x}^{4}-14{x}^{2}+8=0\)
-
\({(x-3)}^{2}-5(x-3)-36=0\)
Tampilkan jawaban
\(x=-1,x=12\)
-
\({(x+2)}^{2}-3(x+2)-54=0\)
-
\({(3y+2)}^{2}+(3y+2)-6=0\)
Tampilkan jawaban
\(x=-\frac{5}{3},x=0\)
-
\({(5y-1)}^{2}+3(5y-1)-28=0\)
-
\({({x}^{2}+1)}^{2}-5({x}^{2}+1)+4=0\)
Tampilkan jawaban
\(x=0,x=\pm \sqrt{3}\)
-
\({({x}^{2}-4)}^{2}-4({x}^{2}-4)+3=0\)
-
\(2{({x}^{2}-5)}^{2}-5({x}^{2}-5)+2=0\)
Tampilkan jawaban
\(x=\pm \frac{\sqrt{22}}{2},x=\pm \sqrt{7}\)
-
\(2{({x}^{2}-5)}^{2}-7({x}^{2}-5)+6=0\)
-
\(x-\sqrt{x}-20=0\)
Tampilkan jawaban
\(x=25\)
-
\(x-8\sqrt{x}+15=0\)
-
\(x+6\sqrt{x}-16=0\)
Tampilkan jawaban
\(x=4\)
-
\(x+4\sqrt{x}-21=0\)
-
\(6x+\sqrt{x}-2=0\)
Tampilkan jawaban
\(x=\frac{1}{4}\)
-
\(6x+\sqrt{x}-1=0\)
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Bagaimana: 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.
Pertanyaan yang sering diajukan
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.
Bagian dari halaman ini diadaptasi dari OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Diringkas dan dijelaskan ulang di sini; kesalahan adalah tanggung jawab kami. Karena sumbernya dilisensikan untuk penggunaan non-komersial, pelajaran ini gratis, menjual apa-apa, dan dibagi di bawah lisensi yang sama.
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