We have f(x) = 2x^2 + 5x -3, and we want to find the values of x for which f(x) > 0.
We start by finding the critical numbers.
Set f(x) = 2x^2 + 5x -3 =0 and solve for x.
(2x-1)(x+ 3) = 0
2x-1 = 0 or x+3 = 0
2x = 1
x = ½ = 0.5 or x = -3
The critical numbers are -3 and 0.5.
Now we check for positive and negative regions.
We have three regions to consider: (a) x < -3; (b) -3 < x < 0.5; and (c) x > 0.5.
Case (a): Let x = -4.
Then f(4) = 2×(-4)^2 + 5(-4) -3 =32 – 20 -3 = 9
f(x) is positive when x < -3.
Case (b): Let x = 0.
Then f(0) = 2×0^2 + 5×0 -3 = 0 + 0 -3 = -3
f(x) is negative when -3 < x < 0.5
Case (c): Let x = 1.
Then f(1) = 2×1^2 + 5×1 -3 = 2 +5 – 3 = 4
f(x) is positive when x > 0.5.
![enter image source here]()
Answer: f(x) is positive when x < -3 and x > 0.5.
