Solve the following inequation: 4x^2 + 3x

Thursday, February 27, 2014

To solve the equation 4x^22+3x-5 >
2.

4x^2+3x-5 > 2

We subtract 2
from both sides:

4x^2+3x-5-2 >
0.

4x^2 +3x-7 > 0.

let f(x) =
4x^2+3x-7).

We find the zero of f(x) by quadratic
formula:

x1, x2 = {-b +or-
sqrt(b^2-4ac)}/2a.

Here a = 4, b= 3 and c=
-7.

Therefore x1 = {-3 +sqrt(3^2-
4*4(-7))}/2*4

x1 = {-3+sqrt121}/8 = {-3+11}/8 =
1.

x2 = (-3-11}/8 = -14/8 =
-7/4.

Therefore f(x) = (4x^2+3x-7) = 4(x+7/4)(x-1) >
0

if x < -3/4 or x > 1, when both factors are of the
same sign and the product is positive.

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