# X^2 + x + 1 > 2

1. Apr 5, 2012

### jrdunaway

1. The problem statement, all variables and given/known data

Solve for X

2. Relevant equations

x^2 + x + 1 > 2

3. The attempt at a solution

x2 + x - 1 > 0

(x + 2)(x - 1/2) > 0

x > 2
x > 1/2

Wolfram Alpha said that the solutions for X are:

x>1/2 (sqrt(5)-1)

and

x<1/2 (-1-sqrt(5))

what did I do wrong?

Last edited: Apr 5, 2012
2. Apr 5, 2012

### Curious3141

Well, for one thing, your factorisation of x2 + x - 1 to (x + 2)(x - 1/2) is wrong.

You can see that by multiplying those factors together. They won't equal the original quadratic.

What's the discriminant of the quadratic?

The other error is in supposing that, if AB > 0, then A > 0 and B > 0 is the only solution. Don't forget that A < 0 and B < 0 is also a valid solution. Remember, if you multiply two negative numbers together, you get a positive.

Another mistake was going from x + 2 > 0 to x > 2. Shouldn't that be x > -2?

The last mistake was in the way you expressed your solution. Even though you considered the A > 0 and B > 0 case, you wrote the solution as x > 2, x > 1/2. Remember the "AND" condition. If x > 2 AND x > 1/2, you should simply "collapse" that solution to x > 2 (the bigger value) since everything that's bigger than 2 is also bigger than 1/2. Of course, 2 and 1/2 are wrong values for this question (as I mentioned earlier), but I'm using them to demonstrate the concept for you.

Last edited: Apr 5, 2012
3. Apr 5, 2012

### D H

Staff Emeritus
Here is what you did wrong:
(x+2)(x-1/2) = x2 + 1.5x - 1, not x2 + x - 1.

4. Apr 5, 2012

### mtayab1994

Why would you go trying to factor when all you have to do is bring the 2 to the left side and solve the equation (x^2+x+?=0) by finding the discriminant Δ and you'll get 2 real roots. Then go on from there.

5. Apr 6, 2012

### verty

Another method is to complete the square, getting p^2 > ...