Solve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation.
[tex]$2 x^2+7 x-4\ \textless \ 0$[/tex]
Answer (1)
Factor the quadratic expression: 2 x 2 + 7 x − 4 = ( 2 x − 1 ) ( x + 4 ) .
Find the roots of the equation ( 2 x − 1 ) ( x + 4 ) = 0 , which are x = − 4 and x = 2 1 .
Test the intervals ( − ∞ , − 4 ) , ( − 4 , 2 1 ) , and ( 2 1 , ∞ ) to find where ( 2 x − 1 ) ( x + 4 ) < 0 .
Express the solution in interval notation: ( − 4 , 2 1 ) .
Explanation
Problem Analysis We are given the quadratic inequality 2 x 2 + 7 x − 4 < 0 . Our goal is to find the solution set for this inequality, express it in interval notation, and graph it on a real number line.
Factoring the Quadratic First, we need to factor the quadratic expression 2 x 2 + 7 x − 4 . We are looking for two numbers that multiply to 2 × − 4 = − 8 and add up to 7 . These numbers are 8 and − 1 . So we can rewrite the middle term as 8 x − x . Thus, we have:
2 x 2 + 8 x − x − 4 < 0
Now, we factor by grouping:
2 x ( x + 4 ) − 1 ( x + 4 ) < 0
( 2 x − 1 ) ( x + 4 ) < 0
Finding the Roots Next, we find the roots of the quadratic equation ( 2 x − 1 ) ( x + 4 ) = 0 . These roots are the critical points that divide the number line into intervals. The roots are:
2 x − 1 = 0 ⟹ x = 2 1
x + 4 = 0 ⟹ x = − 4
Testing the Intervals Now we create a number line and mark the critical points − 4 and 2 1 . These points divide the number line into three intervals: ( − ∞ , − 4 ) , ( − 4 , 2 1 ) , and ( 2 1 , ∞ ) . We need to test a value from each interval to determine where the inequality ( 2 x − 1 ) ( x + 4 ) < 0 is true.
Interval ( − ∞ , − 4 ) : Let's test x = − 5 .
0"> ( 2 ( − 5 ) − 1 ) (( − 5 ) + 4 ) = ( − 10 − 1 ) ( − 1 ) = ( − 11 ) ( − 1 ) = 11 > 0 . So the inequality is not true in this interval.
Interval ( − 4 , 2 1 ) : Let's test x = 0 .
( 2 ( 0 ) − 1 ) ( 0 + 4 ) = ( − 1 ) ( 4 ) = − 4 < 0 . So the inequality is true in this interval.
Interval ( 2 1 , ∞ ) : Let's test x = 1 .
0"> ( 2 ( 1 ) − 1 ) ( 1 + 4 ) = ( 1 ) ( 5 ) = 5 > 0 . So the inequality is not true in this interval.
Expressing the Solution in Interval Notation The inequality ( 2 x − 1 ) ( x + 4 ) < 0 is true in the interval ( − 4 , 2 1 ) . Since the inequality is strict ( < 0 ), we use parentheses to denote that the endpoints are not included in the solution set.
Therefore, the solution set in interval notation is ( − 4 , 2 1 ) .
Final Answer The solution set is the interval ( − 4 , 2 1 ) . This means that all values of x between − 4 and 2 1 (excluding − 4 and 2 1 ) satisfy the inequality 2 x 2 + 7 x − 4 < 0 .
Examples
Understanding quadratic inequalities helps in various real-world scenarios. For instance, a company might want to determine the range of production levels that yield a profit. If the profit equation is a quadratic function, solving a quadratic inequality can help find the production levels where the profit is positive. Similarly, in physics, understanding projectile motion often involves solving quadratic equations or inequalities to determine when an object reaches a certain height or distance.
