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Q. 1

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Found in: Page 505

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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

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

See the step by step solution

## 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 $a{x}^{2}+bx+c=0$.

Comparing it with the given equation we get,

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

Discriminant of the equation is

role="math" localid="1646502307564" ${b}^{2}-4ac\phantom{\rule{0ex}{0ex}}={\left(1\right)}^{2}-4\left(3\right)\left(-1\right)\phantom{\rule{0ex}{0ex}}=1+12\phantom{\rule{0ex}{0ex}}=13>0$

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

## Step 3. Finding the roots

Using the quadratic formula to find the roots:

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

Substituting the values we get,

role="math" localid="1646502325187" $x=\frac{-1±\sqrt{{1}^{2}-4\left(3\right)\left(-1\right)}}{2\left(3\right)}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1±\sqrt{1+12}}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1±\sqrt{13}}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1-\sqrt{13}}{6},\frac{-1+\sqrt{13}}{6}$

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