A little help with an inequality question

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#1
Hi all, I was doing some revision last night and i was looking at this inequality quadratic:

x^2 + 2x + 3 > 0
I end up with (x+1) ^2 > -2, so what is the answer to this?
0
15 years ago
#2
Well any squared term is going to be greater than or equal to zero, and therefore greater than -2, so any x will satisfy it.
0
15 years ago
#3
no real roots because root of b^2 - 4ac is less than 0 (in other words, the answer to your question is x= + or - infinity). Hope this helps
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#4
Hm, can i say this then?

(x+1) ^2 > -2
(x+1)^2 + 2 > 0
(x+1) + sqroot 2 > 0
x > -(1+sqroot2) ?
0
15 years ago
#5
(Original post by slaw)
Hm, can i say this then?

(x+1) ^2 > -2
(x+1)^2 + 2 > 0
(x+1) + sqroot 2 > 0
x > -(1+sqroot2) ?
No.

As JamesF said, any x will satisfy the inequality.
0
15 years ago
#6
(Original post by slaw)
Hi all, I was doing some revision last night and i was looking at this inequality quadratic:

x^2 + 2x + 3 > 0
I end up with (x+1) ^2 > -2, so what is the answer to this?
x^2 + 2x + 3 = (x + 1 - i√2)(x + 1 + i√2) I think.
It has no real roots- the curve never goes below the x-axis.
0
15 years ago
#7
(Original post by slaw)
Hi all, I was doing some revision last night and i was looking at this inequality quadratic:

x^2 + 2x + 3 > 0
I end up with (x+1) ^2 > -2, so what is the answer to this?
If you really need to prove something, I suppose you could always use

f'(x) = 2x+2, giving a turning point at (-1,2)
f''(x) = 2, telling you it's a minimum, so always y=>2

Aitch
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#8
Okay thanks guys
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