# Solve the following inequation: 4x^2 + 3x - 5 > 2

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve 4x^2 + 3x - 5 > 2

4x^2 + 3x - 5 > 2

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

=> 4x^2 + 7x – 4x – 7 > 0

=> x (4x + 7) – 1(4x + 7) >0

=> (x - 1) (4x + 7) > 0

This is possible if either (x - 1) > 0 and (4x + 7) > 0 or (x - 1) < 0 and (4x + 7) < 0.

Now if (x - 1) > 0 and (4x + 7) > 0, we have x > 1 and x > -7/4. All values of x greater than 1 satisfy these relations.

If (x - 1) < 0 and (4x + 7) < 0, we have x < 1 and x < -7/4. All values of x less than -7/4 satisfy these relations.

Therefore the values of x greater than 1 or less than -7/4 satisfy the inequation for 4x^2 + 3x - 5 > 2.

neela | High School Teacher | (Level 3) Valedictorian

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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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll subtract 2:

4x^2 + 3x - 5 - 2 > 2-2

4x^2 + 3x - 7 > 0

The inequality is true if the discriminant of the quadratic is negative:

delta = b^2 - 4ac

We'll identify a,b,c:

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

delta = 9 + 112

delta = 121

x1 = (-3+sqrt121)/8

x1 = (-3+11)/8

x1 = 1

x2 = -14/8

x2 = -7/4

The inequality is positive when x has values in the intervals (-infinity;-7/4) or (1 ; +infinity).