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. $$2(1 + \sqrt{x})^2=13(1 + \sqrt{x}) - 6$$ $$u=(1 + \sqrt{x})$$ $$2u^2=13u - 6$$ $$2u^2 - 13u + 6=0$$ 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. $$2(1 + \sqrt{x})^2=13(1 + \sqrt{x}) - 6$$ $$2(1 + \sqrt{25})^2=13(1 + \sqrt{25}) - 6$$ $$2(1 + 5)^2=13(1 + 5) - 6$$ $$2(6)^2=13(6) - 6$$ $$2(36)=6(13 - 1)$$ $$72=6(12)$$ $$72=72 \hspace{.25em}\color{green}{✔}$$

#### Skills Check:

Example #1

Solve each equation. $$6x^4 - 17x^2 + 7=0$$

Please choose the best answer.

A
$$x=\pm \frac{\sqrt{5}}{2}, \pm \frac{\sqrt{19}}{3}$$
B
$$x=7 \pm 2i, \pm \frac{\sqrt{21}}{3}$$
C
$$x=\pm 3i, \pm \frac{\sqrt{19}}{3}$$
D
$$x=\pm \frac{\sqrt{7}}{2}, \pm \frac{\sqrt{14}}{3}$$
E
$$x=\pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{21}}{3}$$

Example #2

Solve each equation. $$10x^{\frac{2}{3}}- 13x^{\frac{1}{3}}- 9=0$$

Please choose the best answer.

A
$$x=-\frac{1}{7}, \frac{13}{3}$$
B
$$x=3 \pm 2i\sqrt{5}$$
C
$$x=-\frac{1}{4}, \frac{17}{125}$$
D
$$x=-\frac{1}{8}, \frac{729}{125}$$
E
$$x=-\frac{2}{7}, 3$$

Example #3

Solve each equation. $$(3x-1)^2 + 4(3x-1)-2=0$$

Please choose the best answer.

A
$$x=5 \pm 3i$$
B
$$x=-\frac{1}{3}, \frac{5}{7}$$
C
$$x=-\frac{2 \pm \sqrt{5}}{3}$$
D
$$x=-\frac{1 \pm \sqrt{6}}{3}$$
E
$$x=-\frac{7 \pm \sqrt{4}}{3}$$      Congrats, Your Score is 100% Better Luck Next Time, Your Score is %

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