Given the equation 2(x^4) + 4(x^2) + 1 = 0 find all real solutions....?
2014-09-24 7:37 pm
Intuitively, I can see for any x the value of the function will be positive, and at x = 0 the y value is at its minimum, 1. So there are no real solutions, how can I prove this? I tried substitution (let w = x^2) but the discriminant I keep finding to be positive, which I was hoping to find negative.
回答 (2)
2014-09-24 7:41 pm
✔ 最佳答案
Let w = x^2, so we have:2w^2 + 4w + 1 = 0
w = (-4 +/- sqrt(4^2 - 4*2*1)) / (2*2)
w = (-4 +/- sqrt(16 - 8)) / 4
w = (-4 +/- sqrt(8)) / 4
w = -1 +/- sqrt(0.5)
w =~ -0.29289321881345247559915563789515 or -1.7071067811865475244008443621048
x^2 =~ -0.29289321881345247559915563789515 or -1.7071067811865475244008443621048
Since x^2 is negative, then x can't be real.
2014-09-24 7:42 pm
let y = x^2
then y^2 = x^4
2y^2 + 4y + 1 = 0
quad formula
y = (-4 +/- sqrt(8))/4
x^2 = (-4 +/- sqrt(8))/4
x = +/- sqrt((-4 +/- sqrt(8)))/2
-4 - sqrt(8) will be negative, obviously...
4 > sqrt(8), so -4 + sqrt(8) is also negative
you can't sqrt negative numbers and get real solutions... so there are no real solutions to this
http://www.wolframalpha.com/input/?i=2%28x%5E4%29+%2B+4%28x%5E2%29+%2B+1+%3D+0
then y^2 = x^4
2y^2 + 4y + 1 = 0
quad formula
y = (-4 +/- sqrt(8))/4
x^2 = (-4 +/- sqrt(8))/4
x = +/- sqrt((-4 +/- sqrt(8)))/2
-4 - sqrt(8) will be negative, obviously...
4 > sqrt(8), so -4 + sqrt(8) is also negative
you can't sqrt negative numbers and get real solutions... so there are no real solutions to this
http://www.wolframalpha.com/input/?i=2%28x%5E4%29+%2B+4%28x%5E2%29+%2B+1+%3D+0
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https://hk.answers.yahoo.com/question/index?qid=20140924113758AAalkYE