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Q. 1

Expert-verified

Precalculus Enhanced with Graphing Utilities

Found in: Page 505

Book edition 6th

Author(s) Sullivan

Pages 1200 pages

ISBN 9780321795465

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Short Answer

Find the real solutions, if any, of the equation $3 x^{2} + x - 1 = 0$ .

The solutions are $\{\frac{- 1 - \sqrt{13}}{6}, \frac{- 1 + \sqrt{13}}{6}\}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

We are given the equation $3 x^{2} + x - 1 = 0$ and we need to find the real roots of it.

Step 2. Finding the discriminant

The standard form of a quadratic equation is $a x^{2} + b x + c = 0$ .

Comparing it with the given equation we get,

$a = 3, b = 1, c = - 1$ .

Discriminant of the equation is

role="math" localid="1646502307564" $b^{2} - 4 a c = {(1)}^{2} - 4 (3) (- 1) = 1 + 12 = 13 > 0$

Thus there are two unequal real solutions for $x$ .

Step 3. Finding the roots

Using the quadratic formula to find the roots:

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

Substituting the values we get,

role="math" localid="1646502325187" $x = \frac{- 1 \pm \sqrt{1^{2} - 4 (3) (- 1)}}{2 (3)} \Rightarrow x = \frac{- 1 \pm \sqrt{1 + 12}}{6} \Rightarrow x = \frac{- 1 \pm \sqrt{13}}{6} \Rightarrow x = \frac{- 1 - \sqrt{13}}{6}, \frac{- 1 + \sqrt{13}}{6}$

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