Other Types of Equations: Apply It 1

  • Solve equations that include fractions with variables.
  • Solve equations with roots and fractional powers.
  • Use factoring to find solutions to polynomial equations.
  • Find solutions to equations that involve absolute values.
  • Solve absolute value inequalities.

Solving Equations that Resemble Quadratics

Equations that resemble quadratic forms consist of three terms where the first term has a power that differs from [latex]2[/latex]. The exponent of the middle term is half that of the first term, and the third term is a constant. These equations can be treated similarly to standard quadratic equations.

Examples of such equations include [latex]{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0[/latex], and [latex]{x}^{\frac{2}{3}}+4{x}^{\frac{1}{3}}+2=0[/latex].

In these examples, if you multiply the exponent of the middle term by two, you get the exponent of the first term. To solve these equations, we can use a simpler approach by replacing the middle term with a new variable. This makes it easier to solve the equation, just like a regular quadratic equation.

Solve the following equation:

[latex]3{x}^{4}-13{x}^{2}+4=0[/latex]

To solve the equation [latex]3{x}^{4}-13{x}^{2}+4=0[/latex], you can treat it as a quadratic in form by substituting [latex]u = x^2[/latex].

  • Step 1: Substitute [latex]u = x^2[/latex].

    [latex]3{u}^{2}-13u+4=0[/latex]

  • Step 2: Solve the quadratic equation.[latex]\begin{align*} \text{Quadratic equation} & : & 3u^2 - 13u + 4 = 0 \\ \text{Quadratic formula} & : & u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \text{Substitute $a = 3$, $b = -13$, $c = 4$} & : & u = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \\ & : & u = \frac{13 \pm \sqrt{169 - 48}}{6} \\ & : & u = \frac{13 \pm \sqrt{121}}{6} \\ & : & u = \frac{13 \pm 11}{6} \\ \text{Simplify} & : & u = \frac{24}{6} = 4, \quad u = \frac{2}{6} = \frac{1}{3} \end{align*}[/latex]
  • Step 3: Substitute back to find [latex]x[/latex].
    • When [latex]u = 4, x^2 = 4, \text{ so } x = \pm 2.[/latex]
    • When [latex]u = \dfrac{1}{3}, x^2 = \dfrac{1}{3}, \text{ so } x = \pm \dfrac{1}{\sqrt{3}} = \pm \dfrac{\sqrt{3}}{3}.[/latex]

Therefore, the solution to [latex]3{x}^{4}-13{x}^{2}+4=0[/latex] are [latex]x=\pm {2}[/latex] and [latex]\pm \dfrac{\sqrt{3}}{3}[/latex].

Solve the equation: [latex]{\left(x+2\right)}^{2}+11\left(x+2\right)-12=0[/latex]