maths.free › Algebra › Quadratic Equations and Functions › Solve Equations in Quadratic Form
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?
Tani ti. Asnjë kalkulator nuk e zgjidh këtë, por pjesët e saj janë të llogaritura. Provo një më poshtë, ose shkruaj tënde.
Praktiko (40)
Provo secilin në letër së pari. Zbulo përgjigjen për të kontrolluar; ato të verifikuara mund të hapen në zgjidhës për çdo hap.
-
Factor by substitution: \({y}^{4}-{y}^{2}-20.\)
Zbulo përgjigjen
\(\left({y}^{2}+4\right)\left({y}^{2}-5\right)\)
-
Factor by substitution: \({(y-4)}^{2}+8(y-4)+15.\)
Zbulo përgjigjen
\((y-1)(y+1)\)
-
Simplify: ⓐ \({x}^{\frac{1}{2}}\cdot {x}^{\frac{1}{4}}\) ⓑ \({({x}^{\frac{1}{3}})}^{2}\) ⓒ \({({x}^{-1})}^{2}.\)
Zbulo përgjigjen
ⓐ \({x}^{\frac{3}{4}}\); ⓑ \({x}^{\frac{2}{3}}\); ⓒ \({x}^{-2}\)
-
Solve: \(6{x}^{4}-7{x}^{2}+2=0\)
-
Solve: \({x}^{4}-6{x}^{2}+8=0\).
Zbulo përgjigjen
\(x=\sqrt{2},x=\text{-}\sqrt{2},x=2,x=-2\)
-
Solve: \({x}^{4}-11{x}^{2}+28=0\).
Zbulo përgjigjen
\(x=\sqrt{7},x=\text{-}\sqrt{7},x=2,x=-2\)
-
Solve: \({(x-2)}^{2}+7(x-2)+12=0.\)
Zbulo përgjigjen
Prepare for the substitution. Let \(u=x-2\) and substitute. Solve by factoring. Replace \(u\) with \(x-2.\) Solve for \(x.\) Check: -
Solve: \({(x-5)}^{2}+6(x-5)+8=0.\)
Zbulo përgjigjen
\(x=3,x=1\)
-
Solve: \({(y-4)}^{2}+8(y-4)+15=0.\)
Zbulo përgjigjen
\(y=-1,y=1\)
-
Solve: \(x-3\sqrt{x}+2=0.\)
Zbulo përgjigjen
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: \(x-7\sqrt{x}+12=0.\)
Zbulo përgjigjen
\(x=9,x=16\)
-
Solve: \(x-6\sqrt{x}+8=0.\)
Zbulo përgjigjen
\(x=4,x=16\)
-
Solve: \({x}^{\frac{2}{3}}-2{x}^{\frac{1}{3}}-24=0.\)
Zbulo përgjigjen
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: \({x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}-14=0.\)
Zbulo përgjigjen
\(x=-8,x=343\)
-
Solve: \({x}^{\frac{1}{2}}-8{x}^{\frac{1}{4}}+15=0.\)
Zbulo përgjigjen
\(x=81,x=625\)
-
Solve: \(3{x}^{-2}-7{x}^{-1}+2=0.\)
Zbulo përgjigjen
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: \(8{x}^{-2}-10{x}^{-1}+3=0.\)
Zbulo përgjigjen
\(x=\frac{4}{3}x=2\)
-
Solve: \(6{x}^{-2}-23{x}^{-1}+20=0.\)
Zbulo përgjigjen
\(x=\frac{2}{5},x=\frac{3}{4}\)
-
\({x}^{4}-7{x}^{2}+12=0\)
Zbulo përgjigjen
\(x=\pm \sqrt{3},x=\pm 2\)
-
\({x}^{4}-9{x}^{2}+18=0\)
-
\({x}^{4}-13{x}^{2}-30=0\)
Zbulo përgjigjen
\(x=\pm \sqrt{15},x=\pm \sqrt{2}i\)
-
\({x}^{4}+5{x}^{2}-36=0\)
-
\(2{x}^{4}-5{x}^{2}+3=0\)
Zbulo përgjigjen
\(x=\pm 1,x=\frac{\pm \sqrt{6}}{2}\)
-
\(4{x}^{4}-5{x}^{2}+1=0\)
-
\(2{x}^{4}-7{x}^{2}+3=0\)
Zbulo përgjigjen
\(x=\pm \sqrt{3},x=\pm \frac{\sqrt{2}}{2}\)
-
\(3{x}^{4}-14{x}^{2}+8=0\)
-
\({(x-3)}^{2}-5(x-3)-36=0\)
Zbulo përgjigjen
\(x=-1,x=12\)
-
\({(x+2)}^{2}-3(x+2)-54=0\)
-
\({(3y+2)}^{2}+(3y+2)-6=0\)
Zbulo përgjigjen
\(x=-\frac{5}{3},x=0\)
-
\({(5y-1)}^{2}+3(5y-1)-28=0\)
-
\({({x}^{2}+1)}^{2}-5({x}^{2}+1)+4=0\)
Zbulo përgjigjen
\(x=0,x=\pm \sqrt{3}\)
-
\({({x}^{2}-4)}^{2}-4({x}^{2}-4)+3=0\)
-
\(2{({x}^{2}-5)}^{2}-5({x}^{2}-5)+2=0\)
Zbulo përgjigjen
\(x=\pm \frac{\sqrt{22}}{2},x=\pm \sqrt{7}\)
-
\(2{({x}^{2}-5)}^{2}-7({x}^{2}-5)+6=0\)
-
\(x-\sqrt{x}-20=0\)
Zbulo përgjigjen
\(x=25\)
-
\(x-8\sqrt{x}+15=0\)
-
\(x+6\sqrt{x}-16=0\)
Zbulo përgjigjen
\(x=4\)
-
\(x+4\sqrt{x}-21=0\)
-
\(6x+\sqrt{x}-2=0\)
Zbulo përgjigjen
\(x=\frac{1}{4}\)
-
\(6x+\sqrt{x}-1=0\)
Një llogari e lirë shtohet shënime në çdo mësim, një regjistrim të asaj që ju keni përfunduar, problemet tuaja të zgjidhura në një vend, dhe një mësues që ju mund të pyesni rreth kësaj faqeje. Matematika vetë është e hapur për të gjithë, të regjistruar apo jo.
Regjistrohu HyrSimbolet e përdorura këtu
Prek çdo simbol për përkufizimin e plotë, një fotografi dhe se çfarë do të thotë çdo shkronjë në të.
Si: 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.
Pyetja që bëjnë njerëzit
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.
Pjesa e kësaj faqeje është adaptuar nga OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). E përmbledhur dhe ri-shkruar këtu; gabimet janë tona. Për shkak se burimi i tij është licencuar për përdorim jo komercial, ky mësim është i lirë, nuk shet asgjë dhe ndahet nën të njëjtën licencë.
Më shumë në Algebra
Linear equationsQuadratic equationsSystems of equationsInequalitiesFactoringExpandingSimplifying expressionsFunctions and graphsExponential and logarithmic equationsPolynomial equationsAbsolute value