solve for x if `x^2+3x-4 <0`

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`x^2+3x-4`

`= x^2+4x-x-4`

`= x(x+4)-1(x+4)`

`= (x+4)(x-1)`

`x^2+3x-4 < 0`

`(x+4)(x-1) < 0`

`(x+4)(x-1) = 0` when x = -4 and x = 1

When `x<-4` (use x = -5) then `(x+4)(x-1) > 0`

When `1>x>(-4)` (use x = 0) then `(x+4)(x-1) < 0 `

When `x>1` (use x = 2) then `(x+4)(x-1) > 0 `

*So the `(x+4)(x-1) < 0` when `1>x>(-4)` .*

*So the answer is `x in (-4,1)` *

Solve `x^2+3x-4<0` :

Consider the graph of `y=x^2+3x-4=(x+4)(x-1)` :

This is a parabola that opens up with x-intercepts -4 and 1. The parabola must extend below the x-axis (y<0) between -4 and 1.

The solution is the interval -4<x<1 or (-4,1).

The graph:

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