Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To find the endpoints of the solution intervals for the quadratic inequality [tex]\( x^2 + 6x > 16 \)[/tex], we need to treat this as a quadratic equation initially to find the critical points where the quadratic expression equals 16. Follow these steps:
1. Rewrite the Inequality:
Rewrite the inequality as an equality:
[tex]\[ x^2 + 6x = 16 \][/tex]
2. Transform to Standard Form:
Move 16 to the left side to get a standard quadratic equation:
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
3. Solve the Quadratic Equation:
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -16 \)[/tex]:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2} \][/tex]
4. Find the Roots:
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-6 \pm 10}{2} \][/tex]
This gives two solutions:
[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]
5. Check the Intervals:
With the roots [tex]\( x = 2 \)[/tex] and [tex]\( x = -8 \)[/tex], these are the critical points where [tex]\( x^2 + 6x = 16 \)[/tex]. The inequality [tex]\( x^2 + 6x > 16 \)[/tex] requires you to check the intervals determined by these points:
- For [tex]\( x < -8 \)[/tex]
- For [tex]\( -8 < x < 2 \)[/tex]
- For [tex]\( x > 2 \)[/tex]
Since [tex]\( x^2 + 6x \)[/tex] is a parabola opening upwards, [tex]\( x^2 + 6x - 16 \)[/tex] will be negative between the roots and positive outside them. Hence, the solution to the inequality [tex]\( x^2 + 6x > 16 \)[/tex] are the intervals where the expression is positive:
[tex]\[ x \in (-\infty, -8) \cup (2, \infty) \][/tex]
6. Identify the Endpoints:
The endpoints of the solution intervals are [tex]\( x = -8 \)[/tex] and [tex]\( x = 2 \)[/tex].
So, the correct answer is [tex]\( x = -8 \)[/tex] and [tex]\( x = 2 \)[/tex].
1. Rewrite the Inequality:
Rewrite the inequality as an equality:
[tex]\[ x^2 + 6x = 16 \][/tex]
2. Transform to Standard Form:
Move 16 to the left side to get a standard quadratic equation:
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
3. Solve the Quadratic Equation:
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -16 \)[/tex]:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2} \][/tex]
4. Find the Roots:
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-6 \pm 10}{2} \][/tex]
This gives two solutions:
[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]
5. Check the Intervals:
With the roots [tex]\( x = 2 \)[/tex] and [tex]\( x = -8 \)[/tex], these are the critical points where [tex]\( x^2 + 6x = 16 \)[/tex]. The inequality [tex]\( x^2 + 6x > 16 \)[/tex] requires you to check the intervals determined by these points:
- For [tex]\( x < -8 \)[/tex]
- For [tex]\( -8 < x < 2 \)[/tex]
- For [tex]\( x > 2 \)[/tex]
Since [tex]\( x^2 + 6x \)[/tex] is a parabola opening upwards, [tex]\( x^2 + 6x - 16 \)[/tex] will be negative between the roots and positive outside them. Hence, the solution to the inequality [tex]\( x^2 + 6x > 16 \)[/tex] are the intervals where the expression is positive:
[tex]\[ x \in (-\infty, -8) \cup (2, \infty) \][/tex]
6. Identify the Endpoints:
The endpoints of the solution intervals are [tex]\( x = -8 \)[/tex] and [tex]\( x = 2 \)[/tex].
So, the correct answer is [tex]\( x = -8 \)[/tex] and [tex]\( x = 2 \)[/tex].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.