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