Exercise 2 Page 144

We can write a quadratic inequality in one of the following forms, where

a,

b,

and

c

are real numbers and

a \neq = 0 .

0 < a x^{2} + b x + c 0 > a x^{2} + b x + c 0 \leq a x^{2} + b x + c 0 \geq a x^{2} + b x + c

One of the ways to solve a quadratic inequality is by using a graph of the associated function. This is the graphical method. Another way is by using the algebraic method. For this, we need to solve the associated quadratic equation and some test values. We will illustrate how to solve the inequality shown below using both methods.

x^{2} + 6 x - 8 < 0

Solving the Quadratic Inequality by Graphing

Let's start by graphing the corresponding quadratic function. This will allow us to identify the values for which the quadratic inequality holds true.

For this inequality we are just interested in the $x - values$ for which the associated function $f (x) = x^{2} + 6 x - 8$ takes negative values. Let's identify them in the graph.

As we can see, the function goes below the $x - axis$ approximately at $x = - 7.1 .$ Then, it starts going above it at approximately $x = 1.1 .$ Therefore, we can approximate the solution of the inequality set as $- 7.1 < x < 1.1 .$

Solving the Quadratic Inequality by the Algebraic Method

First, we need to write and solve the equation obtained by replacing

<

with

= .

x^{2} + 6 x - 8 < 0 \to x^{2} + 6 x - 8 = 0

We can solve this equation by using the Quadratic Formula.

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

Let's identify the parameters.

a x^{2} + b x + c = 0 (1) x^{2} + 6 x + (- 8) = 0 ⇓ a = 1 b = 6 c = - 8

Now we can proceed by substituting them into the Quadratic Formula to find the solutions.

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

SubstituteValues

Substitute values

x = \frac{- 6 \pm ( 6 ) ^{2} - 4 ( 1 ) ( - 8 )}{2 ( 1 )}

▼

Simplify right-hand side

CalcPow

Calculate power

x = \frac{- 6 \pm 3 6 - 4 ( 1 ) ( - 8 )}{2 ( 1 )}

Multiply

x = \frac{- 6 \pm 3 6 + 3 2}{2}

AddTerms

Add terms

x = \frac{- 6 \pm 6 8}{2}

Rewrite

Rewrite $68$ as $4 \cdot 17$

x = \frac{- 6 \pm 4 t 1 7}{2}

SqrtProd

$a \cdot b = a \cdot b$

x = \frac{- 6 \pm 4 \cdot 1 7}{2}

CalcRoot

Calculate root

x = \frac{- 6 \pm 2 \cdot 1 7}{2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

x = \frac{- 3 \pm 1 7}{1}

DivByOne

$\frac{a}{1} = a$

x = - 3 \pm 17

We can find an approximated form for the solutions using the expression found above.

$x = - 3 \pm 17$
$x = - 3 + 17$	$x = - 3 - 17$
$x = - 3 + 17$	$x = - 3 - 17$
$x = - 3 + 4.123105 \dots$	$x = - 3 - 4.123105 \dots$
$x = 1.23105 \dots$	$x = - 7.123105 \dots$
$x \approx 1.123$	$x \approx - 7.123$

Now we know the critical values of the inequality. We can plot them on a number line using open points. This is because these values do not satisfy the inequality, since the inequality symbol is $< .$ Note that the critical values partition the number line into three intervals.

We can use a test value from each of these intervals to determine whether they satisfy the inequality or not.

	$1^{st}$ Interval	$2^{nd}$ Interval	$3^{rd}$ Interval
Test Value	$x = - 8$	$x = 0$	$x = 2$
Substitute in $x^{2} + 6 x - 8 < 0$	$(- 8)^{2} + 6 (- 8) - 8 < ? 0$	$(0)^{2} + 6 (0) - 8 < ? 0$	$(2)^{2} + 6 (2) - 8 < ? 0$
Is it a solution?	$10 ≮ 0$ $\times$	$- 8 < 0$ $✓$	$8 ≮ 0$ $\times$

Now that we know which intervals belong to the solution set and which of them do not, we can graph the inequality.

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Practice exercises

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Solving the Quadratic Inequality by Graphing

Solving the Quadratic Inequality by the Algebraic Method