# Solve the inequality 2x^2+4x-7<0.

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We have to solve the inequality: 2x^2 + 4x - 7 < 0

2x^2 + 4x - 7 can be written as factors of roots as

(x - x1)(x - x2), where

x1 = -4/4 + sqrt (16 + 56)/4

=> x1 = -1 + 6(sqrt 2)/4

=> x1 = -1 + (3*sqrt 2)/2

x2 = -1 - (3*sqrt 2)/2

As a product of factors we have: (x - (-1 + (3*sqrt 2)/2))(x - (-1 - (3*sqrt 2)/2)

This is negative when either of (x - (-1 + (3*sqrt 2)/2)) or (x - (-1 - (3*sqrt 2)/2) is negative.

- (x - (-1 + (3*sqrt 2)/2)) < 0 and (x - (-1 - (3*sqrt 2)/2) > 0

=> x < (-1 + (3*sqrt 2)/2) and x > (-1 - (3*sqrt 2)/2). This gives the values of x in the interval ( (-1 - (3*sqrt 2)/2), (-1 + (3*sqrt 2)/2).

- (x - (-1 + (3*sqrt 2)/2)) > 0 and (x - (-1 - (3*sqrt 2)/2) < 0

=> x > (-1 + (3*sqrt 2)/2) and x < (-1 - (3*sqrt 2)/2) which is not possible.

**The solution for the inequality is x in the range ((-1 - (3*sqrt 2)/2), (-1 + (3*sqrt 2)/2).**

2x^2 + 4x -7 < 0

First we will find the find the roots.

==> x1= ( -4 + sqrt(16+8*7) / 4

==> x1= ( -4 + 6sqrt2) / 4

==> x1= -1 + 3sqrt2/2

==> x2= -1 -3sqrt2/2

Since the product of the factors should be negative, then the values of x that satisfies the equation should belongs to the interval : (-1-3sqrt2/2 , -1+3sqrt2/2)

**Then the solution for the inequality is the interval x= ( -1-3sqrt2/2 , -1+3sqrt2/2)**

We'll have to calculate the zeroes of the quadratic to verify where the graph of the functin goes below x axis.

To determine the zeroes of quadratic, w'ell apply quadratic formula:

x1 = [-4+sqrt(16 + 56)]/4

x1 = (-4+sqrt72)/4

x1 = (-4+6sqrt2)/4

x1 = (-2+3sqrt2)/2

x1 = -1 + 3sqrt2/2

x2 = -1 - 3sqrt2/2

**The graph of the quadratic function goes below x axis, over the range (-1 - 3sqrt2/2 ; -1 + 3sqrt2/2).**