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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.436. (page 983)

Solve by using the quadratic formula.
$7 y^{2} + 4 y - 3 = 0$

Short Answer

Expert verified

The solution is $y = - 1$ and $y = \frac{3}{7}$ .

Step by step solution

01

Step 1. Given information.

The given equation is $7 y^{2} + 4 y - 3 = 0$ .

02

Step 2. Write the equation in standard form.

The standard form of the quadratic equation is $a x^{2} + b x + c$ .

$7 y^{2} + 4 y - 3 = 7 y^{2} + 4 y + (- 3)$

So, here $a = 7, b = 4, c = - 3$ .

03

Step 3. Apply the quadratic formula.

The quadratic formula is $x = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a}$ .

Substitute the values of $a = 7, b = 4$ and $c = - 3$ in the above formula.

role="math" localid="1645428930014" $\begin{array}{rcl} y & = & \frac{- 4 卤 \sqrt{4^{2} - 4 脳 7 脳 (- 3)}}{2 (7)} \\ = & \frac{- 4 卤 \sqrt{16 + 84}}{14} \\ = & \frac{- 4 卤 \sqrt{100}}{14} \\ = & \frac{- 4 卤 10}{14} \end{array}$

04

Step 4. Now, solve the equation for y.

$\begin{array}{rcl} y & = & \frac{- 4 + 10}{14} \\ = & \frac{6}{14} \\ = & \frac{3}{7} \end{array}$ and $\begin{array}{rcl} y & = & \frac{- 4 - 10}{14} \\ = & \frac{- 14}{14} \\ = & - 1 \end{array}$

05

Step 5. Simplified answer.

Hence, the required solution is $y = \frac{3}{7}$ and $y = - 1$ .

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