How do you solve the inequality x²+x>1. I can not remember how to factor it. It doesnt make sense to me how you can have a single x and a negative 1

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- Jan 15th 2007, 08:29 PMmsdonnellyinequality question
How do you solve the inequality x²+x>1. I can not remember how to factor it. It doesnt make sense to me how you can have a single x and a negative 1

- Jan 15th 2007, 08:32 PMmsdonnellyinequality question
x²+x>1. I can not remmeber how to solve this it doesnt make sense to me how you can have a single x and a negative 1. x(x+1)-1 ???

- Jan 15th 2007, 09:13 PMCaptainBlack
The inequality can be rearranged to:

$\displaystyle x^2+x-1>0$,

but $\displaystyle y=x^2+x-1$ is a parabola opening upwards, so $\displaystyle y>0$ everywhere except between the roots (and at the roots of course)

of $\displaystyle x^2+x-1=0$.

The roots of this quadratic are $\displaystyle \frac{-1 \pm \sqrt{5}}{2}$, so the inequality is satisfied if $\displaystyle x<\frac{-1 - \sqrt{5}}{2}$ or $\displaystyle x>\frac{-1 + \sqrt{5}}{2}$

RonL - Jan 16th 2007, 01:33 AMearboth
- Jan 16th 2007, 04:07 AMtopsquark
$\displaystyle x^2 + x > 1$

$\displaystyle x^2 + x - 1 > 0$

(Take a look at the graph I have attached.)

We are looking for the portion of the x-axis where the graph is above the x-axis.

We do this algebraically by the following process.

Solve $\displaystyle x^2 + x - 1 = 0$

$\displaystyle x = \frac{-1 \pm \sqrt{5}}{2}$ according to the quadratic formula.

Now break the x-axis into three parts:

$\displaystyle (- \infty, \frac{-1 - \sqrt{5}}{2} )$

$\displaystyle ( \frac{-1 - \sqrt{5}}{2} , \frac{-1 + \sqrt{5}}{2} )$

$\displaystyle (\frac{-1 + \sqrt{5}}{2}, \infty )$

We want to test the inequality on each of these intervals. The idea is to pick the easiest number in the interval to do the test on, so for example pick -100000000000000 in the first interval, pick 0 in the second, and pick 100000000000000 in the last interval. (Or any other numbers you like.)

$\displaystyle (- \infty, \frac{-1 - \sqrt{5}}{2} )$: $\displaystyle x^2 + x - 1 > 0$ Check!

$\displaystyle ( \frac{-1 - \sqrt{5}}{2} , \frac{-1 + \sqrt{5}}{2} )$: $\displaystyle x^2 + x - 1 < 0$ Nope!

$\displaystyle (\frac{-1 + \sqrt{5}}{2}, \infty )$: $\displaystyle x^2 + x - 1 > 0$ Check!

So the solution set is $\displaystyle (- \infty, \frac{-1 - \sqrt{5}}{2} ) \cup (\frac{-1 + \sqrt{5}}{2}, \infty )$.

-Dan