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Elementary Algebra

Lynn Marecek, MaryAnne Anthony-Smith

$Math Studyset 91影视 Explanations$ Math

2nd Edition 路 1240 pages

ISBN: 9780998625713

Chapter 10: Q 268. (page 1236)

In the following exercises, solve by using the Quadratic Formula.
$7 y^{2} + 4 y - 3 = 0$

Short Answer

Expert verified

The solutions are $y = - 1 a n d y = \frac{3}{7} .$

Step by step solution

01

Given information.

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

02

Use the quadratic formula.

The quadratic formula for the equation $a x^{2} + b x + c = 0$ is $y = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a}$

Now we compare the given equation with the standard form we get $a = 7, b = 4 a n d c = - 3 .$

Now we substitute these values in the quadratic formula.

localid="1653585740555" $x = \frac{- 4 卤 \sqrt{(4^{2}) - 4 (7) (- 3)}}{2 (7)} y = \frac{- 4 卤 \sqrt{16 + 84}}{14} y = \frac{- 4 卤 \sqrt{100}}{14} y = \frac{- 4 卤 10}{14}$

The two equations arelocalid="1653585775380" $y = \frac{- 4 + 10}{14} a n d y = \frac{- 4 - 10}{14} .$

03

Solve both the equations for x.

$y = \frac{- 4 + 10}{14} y = \frac{6}{14} y = \frac{3}{7} O r y = \frac{- 4 - 10}{14} y = \frac{- 14}{- 14} y = - 1$

Therefore, the two solutions are $y = - 1 a n d y = \frac{3}{7} .$

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