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Find The Solution set of X²-2X-8≥0
Sep 18, 2024
Find The Solution set of X²-2X-8≥0
Solution by Steps
step 1
We start with the inequality X2−2X−8≥0 X^2 - 2X - 8 \geq 0
step 2
To solve this, we first find the roots of the equation X2−2X−8=0 X^2 - 2X - 8 = 0 using the quadratic formula X=−b±b2−4ac2a X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=1,b=−2,c=−8 a = 1, b = -2, c = -8
step 3
Calculating the discriminant: b2−4ac=(−2)2−4(1)(−8)=4+32=36 b^2 - 4ac = (-2)^2 - 4(1)(-8) = 4 + 32 = 36 . Thus, the roots are X=2±62 X = \frac{2 \pm 6}{2}
step 4
This gives us the roots X=4 X = 4 and X=−2 X = -2
step 5
We can now test intervals around the roots: (−∞,−2) (-\infty, -2) , (−2,4) (-2, 4) , and (4,∞) (4, \infty) to determine where the inequality holds
step 6
Testing X=−3 X = -3 in (−∞,−2) (-\infty, -2) : (−3)2−2(−3)−8=9+6−8=7≥0 (-3)^2 - 2(-3) - 8 = 9 + 6 - 8 = 7 \geq 0 (true)
step 7
Testing X=0 X = 0 in (−2,4) (-2, 4) : 02−2(0)−8=−8≥0 0^2 - 2(0) - 8 = -8 \geq 0 (false)
step 8
Testing X=5 X = 5 in (4,∞) (4, \infty) : 52−2(5)−8=25−10−8=7≥0 5^2 - 2(5) - 8 = 25 - 10 - 8 = 7 \geq 0 (true)
step 9
Therefore, the solution set is X≤−2 X \leq -2 or X≥4 X \geq 4
Answer
The solution set is (−∞,−2]∪[4,∞) (-\infty, -2] \cup [4, \infty)
Key Concept
Solving quadratic inequalities involves finding the roots and testing intervals.
Explanation
The solution set includes values of X X that satisfy the inequality X2−2X−8≥0 X^2 - 2X - 8 \geq 0 . The roots divide the number line into intervals, which we test to find where the inequality holds true.
Generate me a similar question
{'context': "[{'Student': 'my name is Jamal, and I love math!', 'Sia': 'that's great!'}, {'Student': 'Solve the inequality 3X² + 12X < 0', 'Sia': ' step 1 ⋮ We start with the inequality \\\ 3X^2 + 12X < 0 \\\\). step 2 ⋮ To solve this, we factor the expression. step 3 ⋮ First, we can factor out a common factor of 3: \\\\( 3(X^2 + 4X) < 0 \\\\. step 4 ⋮ Next, we need to find the roots of the equation \\\\( X^2 + 4X = 0 \\\\. step 5 ⋮ Can you show me how to factor the expression?'}]", 'question': ''}
What are the critical points of the equation x2−2x−8=0x^2 - 2x - 8 = 0?
Generated Graph
Solution by Steps
step 1
To find the critical points of the equation x2−2x−8=0x^2 - 2x - 8 = 0, we first need to solve for xx using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a = 1, b=−2b = -2, and c=−8c = -8
step 2
Calculating the discriminant: b2−4ac=(−2)2−4(1)(−8)=4+32=36b^2 - 4ac = (-2)^2 - 4(1)(-8) = 4 + 32 = 36
step 3
Now, substituting the values into the quadratic formula: x=−(−2)±362(1)=2±62x = \frac{-(-2) \pm \sqrt{36}}{2(1)} = \frac{2 \pm 6}{2}
step 4
This gives us two solutions: x=82=4x = \frac{8}{2} = 4 and x=−42=−2x = \frac{-4}{2} = -2
step 5
Therefore, the critical points of the equation are x=−2x = -2 and x=4x = 4
Answer
The critical points are x=−2x = -2 and x=4x = 4.
Key Concept
Critical points are the values of xx where the function changes its behavior, typically found by solving the equation f(x)=0f(x) = 0.
Explanation
The critical points indicate where the function intersects the x-axis, which is essential for understanding the function's graph and behavior.
Paano Po kukunin Yung True At Saka False?
Para malaman kung ano ang True at False sa isang inequality, kailangan nating tingnan ang mga critical points at ang mga interval na nabuo ng mga ito. 1. Tukuyin ang mga critical points: Sa iyong halimbawa, ang critical points ay nakuha mula sa equation x2−2x−8=0x^2 - 2x - 8 = 0, na nakuha natin sa pamamagitan ng quadratic formula. Ang mga critical points ay x=−2x = -2 at x=4x = 4. 2. Gumawa ng number line: I-plot ang mga critical points sa number line. Ang mga interval na nabuo ay: - (−∞,−2) (-\infty, -2) - (−2,4) (-2, 4) - (4,∞) (4, \infty) 3. Subukan ang mga interval: Pumili ng isang test point mula sa bawat interval at ipasok ito sa original na inequality x2−2x−8≥0x^2 - 2x - 8 \geq 0 upang malaman kung ang inequality ay True o False sa interval na iyon. - Para sa interval (−∞,−2) (-\infty, -2) , subukan ang x=−3x = -3: (−3)2−2(−3)−8=9+6−8=7(True) (-3)^2 - 2(-3) - 8 = 9 + 6 - 8 = 7 \quad (\text{True}) - Para sa interval (−2,4) (-2, 4) , subukan ang x=0x = 0: (0)2−2(0)−8=−8(False) (0)^2 - 2(0) - 8 = -8 \quad (\text{False}) - Para sa interval (4,∞) (4, \infty) , subukan ang x=5x = 5: (5)2−2(5)−8=25−10−8=7(True) (5)^2 - 2(5) - 8 = 25 - 10 - 8 = 7 \quad (\text{True}) 4. Ilagay ang mga resulta: Ang inequality ay True sa intervals (−∞,−2] (-\infty, -2] at [4,∞) [4, \infty) , at False sa interval (−2,4) (-2, 4) . Kaya, ang final na sagot ay: - True: (−∞,−2]∪[4,∞) (-\infty, -2] \cup [4, \infty) - False: (−2,4) (-2, 4) Kung may iba ka pang katanungan, huwag mag-atubiling magtanong!
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