8.3.4 Finding the Solutions

Algebra 18.3.4 Finding the Solutions

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Use what you have learned in the lesson to solve the following quadratic equations.

$5 m^{2} = 80$

Enter the solution to the equation.

Compare your answer:

$m = 4$ and $m = - 4$

$5 m^{2} = 80$

$\frac{5 m^{2}}{5} = \frac{80}{5}$

$m^{2} = 16$

$m = \pm \sqrt{16}$

$m = 4$ and $m = - 4$

$(m - 6)^{2} = 100$

Enter the solution to the equation.

Compare your answer:

$m = 16$ and $m = - 4$

$(m - 6)^{2} = 100$

$(m - 6) = \pm \sqrt{100}$

$m - 6 = 10 m - 6 = - 10$

$m = 16$ and $m = - 4$

$\frac{2}{3} y^{2} - 8 = - 2$

Enter the solution to the equation.

Compare your answer:

$y = 3$ and $y = - 3$

$\begin{matrix} \frac{2}{3} y^{2} - 8 & = & - 2 \\ \frac{2}{3} y^{2} & = & 6 \\ \frac{3}{2} (\frac{2}{3} y^{2}) & = & \frac{3}{2} (6) \\ y^{2} & = & 9 \\ y & = & \pm \sqrt{9} \end{matrix}$ $y = 3$ and $y = - 3$

$\frac{1}{6} (x + 3)^{2} - 8 = 16$

Enter the solution to the equation.

Compare your answer:

$x = 9$ and $x = - 15$

$\begin{matrix} \frac{1}{6} {(x + 3)}^{2} - 8 & = & 16 \\ \frac{1}{6} {(x + 3)}^{2} & = & 24 \\ {(x + 3)}^{2} & = & 144 \\ (x + 3) & = & \pm \sqrt{144} \end{matrix}$ $x + 3 = 12 x + 3 = - 12$ $x = 9$ and $x = - 15$

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- Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis, Jay Abramson, Sharon North, Amy Baldwin, Alyssa Howell
- Publisher/website: OpenStax
- Book title: Algebra 1
- Publication date: Jun 4, 2025
- Location: Houston, Texas
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- Section URL: https://openstax.org/books/algebra-1/pages/8-3-4-finding-the-solutions

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