Lesson Objectives
- Demonstrate an understanding of how to solve quadratic equations
- Learn how to solve an equation that is quadratic in form using substitution
How to Solve Equations that are Quadratic in Form
When a non-quadratic equation can be rewritten as a quadratic equation, we say it is quadratic in form. Suppose we saw the following equation: $$\require{color}x^4-13x^2+36=0$$ Although this equation is not currently a quadratic equation, we can make a substitution and turn it into a quadratic equation.
Let u = x2 $$(x^2)^2 - 13(x^2) + 36=0$$ Plug in a u for x2: $$u^2 - 13u + 36=0$$ Now, we can solve this equation using the quadratic formula.
a = 1, b = -13, c = 36
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{-(-13) \pm \sqrt{(-13)^2 - 4(1)(36)}}{2(1)}$$ $$u=\frac{13 \pm \sqrt{169 - 144}}{2}$$ $$u=\frac{13 \pm \sqrt{25}}{2}$$ $$u=\frac{13 \pm 5}{2}$$ $$u=\frac{13 + 5}{2}=\frac{18}{2}=9$$ $$u=\frac{13 - 5}{2}=\frac{8}{2}=4$$ $$u=4,9$$ Since our original equation was in terms of x, we need to substitute again to solve our equation. Recall that u is x2. $$x^2=4,9$$ $$x^2=4$$ $$x=\pm 2$$ $$x^2=9$$ $$x=\pm 3$$ Our equation has four solutions, x can be -2, 2, -3, or 3.
Let's look at another example.
Example 1: Solve each by substitution. $$10x^{4}+ 7x^{2}- 6=0$$ Let's make a substitution.
Let u = x2 $$10(x^{2})^{2}+ 7x^{2}- 6=0$$ Plug in a u for x2: $$10u^{2}+ 7u - 6=0$$ Now, we can solve this equation using the quadratic formula.
a = 10, b = 7, c = -6
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{-7 \pm \sqrt{(7)^2 - 4(10)(-6)}}{2(10)}$$ $$u=\frac{-7 \pm \sqrt{289}}{20}$$ $$u=\frac{-7 \pm 17}{20}$$ $$u=\frac{-7 + 17}{20}$$ $$u=\frac{10}{20}=\frac{1}{2}$$ $$u=\frac{-7 - 17}{20}$$ $$u=\frac{-24}{20}=-\frac{6}{5}$$ Since our original equation was in terms of x, we need to substitute again to solve our equation. Recall that u is x2. $$x^{2}=-\frac{6}{5}, \frac{1}{2}$$ Let's begin with x2 = -6/5 $$x=\pm \sqrt{-\frac{6}{5}}$$ $$x=\pm i\sqrt{\frac{6}{5}}$$ $$x=\pm \frac{i\sqrt{6}}{\sqrt{5}}$$ $$x=\pm \frac{i\sqrt{30}}{5}$$ Now, let's work on x2 = 1/2 $$x=\pm \sqrt{\frac{1}{2}}$$ $$x=\pm \frac{1}{\sqrt{2}}$$ $$x=\pm \frac{\sqrt{2}}{2}$$ Our solutions for x: $$x=\pm \frac{\sqrt{2}}{2}, \pm \frac{i\sqrt{30}}{5}$$ Let's look at another example.
Example 2: Solve each by substitution. $$2(1 + \sqrt{x})^2=13(1 + \sqrt{x}) - 6$$ $$u=(1 + \sqrt{x})$$ $$2u^2=13u - 6$$ $$2u^2 - 13u + 6=0$$ Now, we can solve this equation using the quadratic formula.
a = 2, b = -13, c = 6
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{-(-13) \pm \sqrt{(-13)^2 - 4(2)(6)}}{2(2)}$$ $$u=\frac{13 \pm \sqrt{169 - 48}}{4}$$ $$u=\frac{13 \pm \sqrt{121}}{4}$$ $$u=\frac{13 \pm 11}{4}$$ $$u=\frac{13 + 11}{4}=\frac{24}{4}=6$$ $$u=\frac{13 - 11}{4}=\frac{2}{4}=\frac{1}{2}$$ $$u=6, \frac{1}{2}$$ Since our original equation was in terms of x, we need to substitute again to solve our equation. Recall that u is 1 plus the square root of x. $$u=6$$ $$1 + \sqrt{x}=6$$ $$\sqrt{x}=5$$ $$x=25$$ $$u=\frac{1}{2}$$ $$1 + \sqrt{x}=\frac{1}{2}$$ $$\sqrt{x}=-\frac{1}{2}$$ We know the principal square root of a number is non-negative. Therefore, there is no solution to this part.
Since we used the squaring property of equality, we need to check our answer (x = 25) in the original equation. $$2(1 + \sqrt{x})^2=13(1 + \sqrt{x}) - 6$$ $$2(1 + \sqrt{25})^2=13(1 + \sqrt{25}) - 6$$ $$2(1 + 5)^2=13(1 + 5) - 6$$ $$2(6)^2=13(6) - 6$$ $$2(36)=6(13 - 1)$$ $$72=6(12)$$ $$72=72 \hspace{.25em}\color{green}{✔}$$ Let's look at one final example.
Example 3: Solve each by substitution. $$2\left(x - \frac{1}{2}\right)^{2}+ 5\left(x - \frac{1}{2}\right) - 3=0$$ Let's make a substitution. $$\text{Let}\hspace{.2em}u=x - \frac{1}{2}$$ Our equation becomes: $$2u^{2}+ 5u - 3=0$$ Now, we can solve this equation using the quadratic formula.
a = 2, b = 5, c = -3
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{(-5) \pm \sqrt{(5)^2 - 4(2)(-3)}}{2(2)}$$ $$u=\frac{-5 \pm 7}{4}$$ $$\frac{-5 + 7}{4}=\frac{1}{2}$$ $$\frac{-5 - 7}{4}=-3$$ $$u=-3, \frac{1}{2}$$ Let's substitute to get our answer in terms of x. We will need to solve two equations. $$x - \frac{1}{2}=-3$$ $$x=-\frac{6}{2}+ \frac{1}{2}$$ $$x=-\frac{5}{2}$$ $$x - \frac{1}{2}=\frac{1}{2}$$ $$x=1$$ $$x=-\frac{5}{2}, 1$$
Let u = x2 $$(x^2)^2 - 13(x^2) + 36=0$$ Plug in a u for x2: $$u^2 - 13u + 36=0$$ Now, we can solve this equation using the quadratic formula.
a = 1, b = -13, c = 36
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{-(-13) \pm \sqrt{(-13)^2 - 4(1)(36)}}{2(1)}$$ $$u=\frac{13 \pm \sqrt{169 - 144}}{2}$$ $$u=\frac{13 \pm \sqrt{25}}{2}$$ $$u=\frac{13 \pm 5}{2}$$ $$u=\frac{13 + 5}{2}=\frac{18}{2}=9$$ $$u=\frac{13 - 5}{2}=\frac{8}{2}=4$$ $$u=4,9$$ Since our original equation was in terms of x, we need to substitute again to solve our equation. Recall that u is x2. $$x^2=4,9$$ $$x^2=4$$ $$x=\pm 2$$ $$x^2=9$$ $$x=\pm 3$$ Our equation has four solutions, x can be -2, 2, -3, or 3.
Let's look at another example.
Example 1: Solve each by substitution. $$10x^{4}+ 7x^{2}- 6=0$$ Let's make a substitution.
Let u = x2 $$10(x^{2})^{2}+ 7x^{2}- 6=0$$ Plug in a u for x2: $$10u^{2}+ 7u - 6=0$$ Now, we can solve this equation using the quadratic formula.
a = 10, b = 7, c = -6
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{-7 \pm \sqrt{(7)^2 - 4(10)(-6)}}{2(10)}$$ $$u=\frac{-7 \pm \sqrt{289}}{20}$$ $$u=\frac{-7 \pm 17}{20}$$ $$u=\frac{-7 + 17}{20}$$ $$u=\frac{10}{20}=\frac{1}{2}$$ $$u=\frac{-7 - 17}{20}$$ $$u=\frac{-24}{20}=-\frac{6}{5}$$ Since our original equation was in terms of x, we need to substitute again to solve our equation. Recall that u is x2. $$x^{2}=-\frac{6}{5}, \frac{1}{2}$$ Let's begin with x2 = -6/5 $$x=\pm \sqrt{-\frac{6}{5}}$$ $$x=\pm i\sqrt{\frac{6}{5}}$$ $$x=\pm \frac{i\sqrt{6}}{\sqrt{5}}$$ $$x=\pm \frac{i\sqrt{30}}{5}$$ Now, let's work on x2 = 1/2 $$x=\pm \sqrt{\frac{1}{2}}$$ $$x=\pm \frac{1}{\sqrt{2}}$$ $$x=\pm \frac{\sqrt{2}}{2}$$ Our solutions for x: $$x=\pm \frac{\sqrt{2}}{2}, \pm \frac{i\sqrt{30}}{5}$$ Let's look at another example.
Example 2: Solve each by substitution. $$2(1 + \sqrt{x})^2=13(1 + \sqrt{x}) - 6$$ $$u=(1 + \sqrt{x})$$ $$2u^2=13u - 6$$ $$2u^2 - 13u + 6=0$$ Now, we can solve this equation using the quadratic formula.
a = 2, b = -13, c = 6
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{-(-13) \pm \sqrt{(-13)^2 - 4(2)(6)}}{2(2)}$$ $$u=\frac{13 \pm \sqrt{169 - 48}}{4}$$ $$u=\frac{13 \pm \sqrt{121}}{4}$$ $$u=\frac{13 \pm 11}{4}$$ $$u=\frac{13 + 11}{4}=\frac{24}{4}=6$$ $$u=\frac{13 - 11}{4}=\frac{2}{4}=\frac{1}{2}$$ $$u=6, \frac{1}{2}$$ Since our original equation was in terms of x, we need to substitute again to solve our equation. Recall that u is 1 plus the square root of x. $$u=6$$ $$1 + \sqrt{x}=6$$ $$\sqrt{x}=5$$ $$x=25$$ $$u=\frac{1}{2}$$ $$1 + \sqrt{x}=\frac{1}{2}$$ $$\sqrt{x}=-\frac{1}{2}$$ We know the principal square root of a number is non-negative. Therefore, there is no solution to this part.
Since we used the squaring property of equality, we need to check our answer (x = 25) in the original equation. $$2(1 + \sqrt{x})^2=13(1 + \sqrt{x}) - 6$$ $$2(1 + \sqrt{25})^2=13(1 + \sqrt{25}) - 6$$ $$2(1 + 5)^2=13(1 + 5) - 6$$ $$2(6)^2=13(6) - 6$$ $$2(36)=6(13 - 1)$$ $$72=6(12)$$ $$72=72 \hspace{.25em}\color{green}{✔}$$ Let's look at one final example.
Example 3: Solve each by substitution. $$2\left(x - \frac{1}{2}\right)^{2}+ 5\left(x - \frac{1}{2}\right) - 3=0$$ Let's make a substitution. $$\text{Let}\hspace{.2em}u=x - \frac{1}{2}$$ Our equation becomes: $$2u^{2}+ 5u - 3=0$$ Now, we can solve this equation using the quadratic formula.
a = 2, b = 5, c = -3
Plug into the quadratic formula: $$u=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$u=\frac{(-5) \pm \sqrt{(5)^2 - 4(2)(-3)}}{2(2)}$$ $$u=\frac{-5 \pm 7}{4}$$ $$\frac{-5 + 7}{4}=\frac{1}{2}$$ $$\frac{-5 - 7}{4}=-3$$ $$u=-3, \frac{1}{2}$$ Let's substitute to get our answer in terms of x. We will need to solve two equations. $$x - \frac{1}{2}=-3$$ $$x=-\frac{6}{2}+ \frac{1}{2}$$ $$x=-\frac{5}{2}$$ $$x - \frac{1}{2}=\frac{1}{2}$$ $$x=1$$ $$x=-\frac{5}{2}, 1$$
Skills Check:
Example #1
Solve each equation. $$12x^{4}- 5x^{2}- 2=0$$
Please choose the best answer.
A
$$x=\pm \frac{i}{3}, \pm \frac{\sqrt{6}}{5}$$
B
$$x=\pm \frac{i}{4}, \pm \frac{\sqrt{6}}{3}$$
C
$$x=\pm 2i, \pm \frac{\sqrt{3}}{5}$$
D
$$x=1 \pm i\sqrt{5}$$
E
$$x=\pm \frac{i}{2}, \pm \frac{\sqrt{6}}{3}$$
Example #2
Solve each equation. $$\left(x - \frac{1}{3}\right)^{2}+ 6\left(x - \frac{1}{3}\right) - 7=0$$
Please choose the best answer.
A
$$x=-2, 7$$
B
$$x=-\frac{20}{3}, \frac{4}{3}$$
C
$$x=-\frac{18}{5}, 3$$
D
$$x=-\frac{2}{7}, \frac{1}{3}$$
E
$$x=-9, 4$$
Example #3
Solve each equation. $$5\left(1 - \sqrt{x}\right)^{2}- 6\left(1 - \sqrt{x}\right) - 11=0$$
Please choose the best answer.
A
$$x=-8, 4$$
B
$$x=4$$
C
$$x=-\frac{7}{4}, 3$$
D
$$x=-\frac{12}{13}, 7$$
E
$$x=-\frac{5}{11}, 12$$
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