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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 9: Q. 140 (page 898)

In the following exercises, solve by using the Quadratic Formula.
$\frac{1}{9} c^{2} + \frac{2}{3} c = 3$

Short Answer

Expert verified

The solutions of the given equation are $b = 3$ and $b = - 9$ .

Step by step solution

01

Step 1. Write the quadratic equation in standard form.

The given equation is:

$\frac{1}{9} c^{2} + \frac{2}{3} c = 3$

The given equation can be written as:

$\begin{array}{rcl} \frac{1}{9} c^{2} + \frac{2}{3} c - 3 & = & 0 \\ \frac{c^{2} + 6 c - 27}{9} & = & 0 \\ c^{2} + 6 c - 27 & = & 0 \end{array}$

02

Step 2. Substitute the values in Quadratic Formula.

Quadratic Formula for the equation $a x^{2} + b x + c = 0$ is:

$x = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a}$

In the given equation, $a = 1, b = 6, c = - 27$ . Substitute these values in Quadratic Formula.

$b = \frac{- 6 卤 \sqrt{(6)^{2} - 4 (1) (- 27)}}{2 (1)}$

03

Step 3. Simplify the value.

On simplification, we get

$\begin{array}{rcl} b & = & \frac{- 6 卤 \sqrt{36 + 108}}{2} \\ b & = & \frac{- 6 卤 \sqrt{144}}{2} \\ b & = & \frac{- 6}{2} 卤 \frac{12}{2} \\ b & = & - 3 卤 6 \end{array}$

It can be wriitten as:

$\begin{array}{rcl} b & = & - 3 + 6 and b = - 3 - 6 \\ b & = & 3 and b = - 9 \end{array}$

04

Step 4. Conclusion.

The solutions of the given equation are $b = 3$ and $b = - 9$ .

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