Solve the following inequalities:
(a) [tex]x^2+6 x+8\ \textless \ 0[/tex]
(b) [tex]4 x^2+x-3\ \textless \ 0[/tex]
(c) [tex]6 x^2+7 x-3\ \textless \ 0[/tex]
(d) [tex]x^2+5 x-14 \leq 0[/tex]

Factor each quadratic inequality.
Find the roots of the corresponding quadratic equation.
Determine the intervals where the inequality holds based on the roots.
Express the solution for each inequality as an interval: (a) − 4 < x < − 2 , (b) − 1 < x < 4 3 ​ , (c) − 2 3 ​ < x < 3 1 ​ , (d) − 7 ≤ x ≤ 2 .

Explanation

Problem Analysis We are given four quadratic inequalities to solve. Our goal is to find the solution set for each inequality. We will factor each quadratic expression and determine the intervals where the inequality holds.

Solving Inequality (a) (a) x 2 + 6 x + 8 < 0 We factor the quadratic expression: x 2 + 6 x + 8 = ( x + 4 ) ( x + 2 ) .
The inequality becomes ( x + 4 ) ( x + 2 ) < 0 .
The roots are x = − 4 and x = − 2 .
We test the intervals ( − ∞ , − 4 ) , ( − 4 , − 2 ) , and ( − 2 , ∞ ) .



For x < − 4 , both ( x + 4 ) and ( x + 2 ) are negative, so their product is positive.
For − 4 < x < − 2 , ( x + 4 ) is positive and ( x + 2 ) is negative, so their product is negative.
For -2"> x > − 2 , both ( x + 4 ) and ( x + 2 ) are positive, so their product is positive. Thus, the solution is − 4 < x < − 2 .


Solving Inequality (b) (b) 4 x 2 + x − 3 < 0 We factor the quadratic expression: 4 x 2 + x − 3 = ( 4 x − 3 ) ( x + 1 ) .
The inequality becomes ( 4 x − 3 ) ( x + 1 ) < 0 .
The roots are x = − 1 and x = 4 3 ​ .
We test the intervals ( − ∞ , − 1 ) , ( − 1 , 4 3 ​ ) , and ( 4 3 ​ , ∞ ) .


For x < − 1 , both ( 4 x − 3 ) and ( x + 1 ) are negative, so their product is positive.
For − 1 < x < 4 3 ​ , ( 4 x − 3 ) is negative and ( x + 1 ) is positive, so their product is negative.
For \frac{3}{4}"> x > 4 3 ​ , both ( 4 x − 3 ) and ( x + 1 ) are positive, so their product is positive. Thus, the solution is − 1 < x < 4 3 ​ .


Solving Inequality (c) (c) 6 x 2 + 7 x − 3 < 0 We factor the quadratic expression: 6 x 2 + 7 x − 3 = ( 2 x + 3 ) ( 3 x − 1 ) .
The inequality becomes ( 2 x + 3 ) ( 3 x − 1 ) < 0 .
The roots are x = − 2 3 ​ and x = 3 1 ​ .
We test the intervals ( − ∞ , − 2 3 ​ ) , ( − 2 3 ​ , 3 1 ​ ) , and ( 3 1 ​ , ∞ ) .


For x < − 2 3 ​ , both ( 2 x + 3 ) and ( 3 x − 1 ) are negative, so their product is positive.
For − 2 3 ​ < x < 3 1 ​ , ( 2 x + 3 ) is positive and ( 3 x − 1 ) is negative, so their product is negative.
For \frac{1}{3}"> x > 3 1 ​ , both ( 2 x + 3 ) and ( 3 x − 1 ) are positive, so their product is positive. Thus, the solution is − 2 3 ​ < x < 3 1 ​ .


Solving Inequality (d) (d) x 2 + 5 x − 14 ≤ 0 We factor the quadratic expression: x 2 + 5 x − 14 = ( x + 7 ) ( x − 2 ) .
The inequality becomes ( x + 7 ) ( x − 2 ) ≤ 0 .
The roots are x = − 7 and x = 2 .
We test the intervals ( − ∞ , − 7 ) , ( − 7 , 2 ) , and ( 2 , ∞ ) .


For x < − 7 , both ( x + 7 ) and ( x − 2 ) are negative, so their product is positive.
For − 7 < x < 2 , ( x + 7 ) is positive and ( x − 2 ) is negative, so their product is negative.
For 2"> x > 2 , both ( x + 7 ) and ( x − 2 ) are positive, so their product is positive. Since the inequality is non-strict, we include the roots in the solution. Thus, the solution is − 7 ≤ x ≤ 2 .


Final Answer The solutions to the inequalities are: (a) − 4 < x < − 2 (b) − 1 < x < 4 3 ​ (c) − 2 3 ​ < x < 3 1 ​ (d) − 7 ≤ x ≤ 2

Examples
Quadratic inequalities are used in various real-world applications, such as determining the range of values for a projectile's trajectory to ensure it lands within a specified area, optimizing the dimensions of a garden to maximize its area while staying within a budget for fencing, or modeling the profit margin of a business based on production costs and sales revenue. Understanding how to solve these inequalities allows us to make informed decisions and predictions in these scenarios.

Answered by GinnyAnswer | 2025-07-03

The inequalities were solved by factoring each quadratic expression and determining the intervals where they are less than or equal to zero. The solutions are: (a) -4 < x < -2, (b) -1 < x < 3/4, (c) -3/2 < x < 1/3, (d) -7 ≤ x ≤ 2. This systematic approach involves finding roots and testing intervals to determine where the inequalities hold true.
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Answered by Anonymous | 2025-07-04