# Math Help - Quadratic Inequality

Hello, I'm new to this forum but I thought it may be helpful to come here for responses when a teacher/classmate is unavailable. I'm currently studying AS level mathematics... Ok so for the homework I have been solving quadratic inequalities by graphing them but this one question out of 10 is a problem for me. The inequality is: x^2 + 2x -1 < 0 I tried factorising it like all the others but it won't factorise so I don't know what coordinates to use to plot it on a graph and therefore can't find the solution. I also tried completing the square to get (x + 1)^2 -2 < 0 and this told me that the line of symmetry of the curve is at x = -1 and that the turning point is at -2 but this didn't help either. Can anyone explain how to do this? Any response will be appreciated.

2. The simplest way to solve an inequality like this, $x^2+ 2x- 1< 0$, or one more complicated, is to solve the corresponding equation, $x^2+ 2x- 1= 0$. That can be solved by completing the square or by using the quadratic formula.
$x= \frac{-3\pm\sqrt{4+4}}{2}= \frac{-3\pm 2\sqrt{2}}{2}$

The point is that a continuous function, such as this, can change from "> 0" to "< 0" only where it is equal to 0.

Those two points, $x= \frac{-3- 2\sqrt{2}}{2}$ and $x= \frac{-3+ 2\sqrt{2}}{2}$ divides the number line into three intervals:
$x< \frac{-3- 2\sqrt{2}+}{2}$
$\frac{-3- 2\sqrt{2}+}{2}< x< \frac{-3+ 2\sqrt{2}+}{2}$
$x> \frac{-3+ 2\sqrt{2}+}{2}$

Choose one point in each interval: since [tex]2\sqrt{2}[/itex] is between 2 and 3, $\frac{-3- 2\sqrt{2}+}{2}$ is between -5/2 and 1/2. -3 is in the first interval and 0 is in the second.
$(-3)^2+ 2(-3)-1= 9- 6- 1= 2> 0$ so every $x< \frac{-1- 2\sqrt{2}}{2}$ does not satisify the inequality.

$0^2+ 2(0)-1= -1< 0$ so every $\frac{-3- 2\sqrt{2}+}{2}< x< \frac{-3+ 2\sqrt{2}+}{2}$ satifies the inequality.

x= 2 is in the last interval and $2^2+ 2(2)- 1= 8- 1= 7> 0$ so any $x> \frac{-3+ 2\sqrt{2}+}{2}$ does not satisfy the inequality.

The inequality is satisfied for $\frac{-3- 2\sqrt{2}+}{2}< x< \frac{-3+ 2\sqrt{2}+}{2}$.

Of course, for this simple inequality, that is exactly the same as saying that the graph of [tex]y= x^2+ 2x- 1[/itex] is a parabola opening upward and so y< 0 for x between the two x-intercepts.