+0

# x^2+x>1/e^4

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x^2+x>1/e^4

Guest Dec 26, 2015

#3
+78719
+10

x^2+x >1/e^4        make this an equality and complete the square

x^2 + x + 1/4 = 1/e^4 + 1/4       factor the left side, simplify the right

(x + 1/2)^2  = [ 4 + e^4] / [4e^4)   take the square root of both sides

x + 1/2  =  +/- sqrt [ 4 + e^4] / [2e^2]

x = +/- sqrt [ 4 + e^4] / [ 2e^2] - 1/2  ≈  [ -1.0179919293664084, 0.0179919293664084 ]

The answer  will come from one or more of these intervals

(-infinity, - sqrt [ 4 + e^4] / [ 2e^2] - 1/2) , (- sqrt [ 4 + e^4] / [ 2e^2] - 1/2,  sqrt [ 4 + e^4] / [ 2e^2] - 1/2), ( sqrt [ 4 + e^4] / [ 2e^2] - 1/2, infinity)

Choosing 0 as a test point for the middle interval will make the inequality false

So.....the two outside intervals solve the inequality

Here's the graph that proves this........https://www.desmos.com/calculator/iluvf1vudp

CPhill  Dec 26, 2015
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#1
0

x^2+x>1/e^4

=x>sqrt(4+e^4)/(2e^2)-1/2

Guest Dec 26, 2015
#2
+5

https://www.symbolab.com

Guest Dec 26, 2015
#3
+78719
+10

x^2+x >1/e^4        make this an equality and complete the square

x^2 + x + 1/4 = 1/e^4 + 1/4       factor the left side, simplify the right

(x + 1/2)^2  = [ 4 + e^4] / [4e^4)   take the square root of both sides

x + 1/2  =  +/- sqrt [ 4 + e^4] / [2e^2]

x = +/- sqrt [ 4 + e^4] / [ 2e^2] - 1/2  ≈  [ -1.0179919293664084, 0.0179919293664084 ]

The answer  will come from one or more of these intervals

(-infinity, - sqrt [ 4 + e^4] / [ 2e^2] - 1/2) , (- sqrt [ 4 + e^4] / [ 2e^2] - 1/2,  sqrt [ 4 + e^4] / [ 2e^2] - 1/2), ( sqrt [ 4 + e^4] / [ 2e^2] - 1/2, infinity)

Choosing 0 as a test point for the middle interval will make the inequality false

So.....the two outside intervals solve the inequality

Here's the graph that proves this........https://www.desmos.com/calculator/iluvf1vudp

CPhill  Dec 26, 2015

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