0 ≥ 4 x^2 + 8 x - 5
0 ≥ (2 x + 5) (2 x - 1)
for their product to be positive (>0), both factors must have the same sign
both ≥ 0 ... x ≥ 1/2
both ≤ 0 ... x ≤ -5/2
Solve give your answer in interval notation: 5-4x^2>=8x!!! Please help ASAP!!!:(
2 answers
5-4x^2 ≥ 8x
-4x^2 - 8x + 5 ≥ 0
4x^2 + 8x - 5 ≤ 0
(2x + 5)(2x - 1) ≤ 0
Where is the parabola y = (2x+5)(2x-1) below the x-axis ?
-5/2 ≤ x ≤ 1/2
https://www.wolframalpha.com/input/?i=5-4x%5E2-+8x%E2%89%A50
-4x^2 - 8x + 5 ≥ 0
4x^2 + 8x - 5 ≤ 0
(2x + 5)(2x - 1) ≤ 0
Where is the parabola y = (2x+5)(2x-1) below the x-axis ?
-5/2 ≤ x ≤ 1/2
https://www.wolframalpha.com/input/?i=5-4x%5E2-+8x%E2%89%A50
1 answer