body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Glencoe Algebra 2, 2012

MH

McGraw Hill Glencoe Algebra 2, 2012 View details

8. Quadratic Inequalities

Continue to next subchapter

Exercise 53 Page 287

To solve the quadratic inequality algebraically, we will follow three steps.

Solve the related quadratic equation.
Plot the solutions on a number line.
Test a value from each interval to see if it satisfies the original inequality.

Step $1$

We will start by solving the related equation.

- 3 x^{2} + 10 x = 5 \Leftrightarrow - 3 x^{2} + 10 x + (- 5) = 0

We see above that

a = - 3,

b = 10,

and

c = - 5 .

Let's substitute these values into the Quadratic Formula to solve the equation.

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

SubstituteValues

Substitute values

x = \frac{- 1 0 \pm 1 0 ^{2} - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

▼

Simplify right-hand side

Calculate power

x = \frac{- 1 0 \pm 1 0 0 - 4 ( - 3 ) ( - 5 )}{2 ( - 3 )}

Multiply

x = \frac{- 1 0 \pm 1 0 0 - 6 0}{- 6}

Subtract term

x = \frac{- 1 0 \pm 4 0}{- 6}

Now we can calculate the first root using the positive sign and the second root using the negative sign.

$x = \frac{- 1 0 \pm 4 0}{- 6}$
$x = \frac{- 1 0 + 4 0}{- 6}$	$x = \frac{- 1 0 - 4 0}{- 6}$
$x = \frac{- 1 0}{- 6} + \frac{4 0}{- 6}$	$x = \frac{- 1 0}{- 6} - \frac{4 0}{- 6}$
$x = \frac{1 0}{6} - \frac{4 0}{6}$	$x = \frac{1 0}{6} + \frac{4 0}{6}$
$x \approx 0.61$	$x \approx 2.72$

Step $2$

The solutions of the related equation are approximately $0.61$ and $2.72 .$ Let's plot them on a number line. Since the original is a strict inequality, the points will be open.

Step $3$

Finally, we must test a value from each interval to see if it satisfies the original inequality. Let's choose a value from the first interval,

x < 0.61 .

For simplicity, we will choose

x = 0 .

- 3 x^{2} + 10 x < 5

$x = 0$

- 3 (0)^{2} + 10 (0) < ? 5

▼

Simplify left-hand side

Calculate power

- 3 (0) + 10 (0) < ? 5

Multiply

0 + 0 < ? 5

Identity Property of Addition

0 < 5 ✓

Since

x = 0

produced a true statement, the interval

x < - 0.61

is a part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is It Part of the Solution?
$0.61 < x < 2.72$	$1$	$7 ≮ 5$ $\times$	No
$x > 2.72$	$3$	$3 < 5$ $✓$	Yes

We can now write the solution set and show it on a number line.

{x ∣ x < 0.61 or x > 2.72} or (- \infty, 0.61) \cup (2.72, \infty)

Subchapter links

Check Your Understanding p.285

Practice and Problem Solving p.286-287

H.O.T. Problems p.287

Standardized Test Practice p.288

Spiral Review p.288

Skills Review p.288