Solve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. [tex]x^2-9 x+14\ \textgreater \ 0[/tex] Solve the inequality. What is the solution set? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
Answer (2)
Factor the quadratic expression: x 2 − 9 x + 14 = ( x − 2 ) ( x − 7 ) .
Find the critical points: x = 2 and x = 7 .
Test intervals ( − ∞ , 2 ) , ( 2 , 7 ) , and ( 7 , ∞ ) to determine where 0"> ( x − 2 ) ( x − 7 ) > 0 .
Express the solution set in interval notation: ( − ∞ , 2 ) ∪ ( 7 , ∞ ) .
Explanation
Factor the quadratic expression First, we need to factor the quadratic expression x 2 − 9 x + 14 . We are looking for two numbers that multiply to 14 and add up to -9. These numbers are -2 and -7.
Rewrite the inequality So, we can rewrite the quadratic expression as 0"> ( x − 2 ) ( x − 7 ) > 0 .
Find the critical points Now, we need to find the critical points by setting each factor equal to zero: x − 2 = 0 ⇒ x = 2 x − 7 = 0 ⇒ x = 7
Determine the intervals These critical points divide the number line into three intervals: ( − ∞ , 2 ) , ( 2 , 7 ) , and ( 7 , ∞ ) . We will test a value from each interval to determine where the inequality 0"> ( x − 2 ) ( x − 7 ) > 0 is true.
Test values in each interval
Interval ( − ∞ , 2 ) : Let's test x = 0 .
0"> ( 0 − 2 ) ( 0 − 7 ) = ( − 2 ) ( − 7 ) = 14 > 0 . So, the inequality is true in this interval.
Interval ( 2 , 7 ) : Let's test x = 3 .
( 3 − 2 ) ( 3 − 7 ) = ( 1 ) ( − 4 ) = − 4 < 0 . So, the inequality is false in this interval.
Interval ( 7 , ∞ ) : Let's test x = 8 .
0"> ( 8 − 2 ) ( 8 − 7 ) = ( 6 ) ( 1 ) = 6 > 0 . So, the inequality is true in this interval.
Express the solution set in interval notation Therefore, the solution set is ( − ∞ , 2 ) ∪ ( 7 , ∞ ) .
Examples
Polynomial inequalities are useful in various real-world scenarios. For example, a company might use them to model the profit margin of a product. Suppose the profit P ( x ) from selling x units of a product is given by P ( x ) = x 2 − 9 x + 14 . The company wants to know for what number of units sold the profit is positive, i.e., 0"> P ( x ) > 0 . Solving the inequality 0"> x 2 − 9 x + 14 > 0 gives the range of units that need to be sold to ensure a positive profit. This helps in making informed business decisions regarding production and sales targets.
The solution set for the inequality 0"> x 2 − 9 x + 14 > 0 is ( − ∞ , 2 ) ∪ ( 7 , ∞ ) . This indicates that the inequality holds true for values of x less than 2 and greater than 7. Critical points were found by factoring and testing intervals between those points.
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