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Prove that the equation ( (x^4) + (3(x^2)) - 2 ) = 0 has exactly 2 real roots.

2006-11-02 08:03:37 · 2 answers · asked by James B 1 in Science & Mathematics Mathematics

2 answers

Let y = x^2

So you have y^2 + 3y - 2
Your discriminant (b² - 4ac) = 3² - 4*1*(-2) = 17

So that means you have two real roots for y^2
y = x^2 = (-3 +/- sqrt(17))/2
y = x^2 = 0.561552813 or -3.56155281

When you solve for x, you have:
x = sqrt(0.561552813) or sqrt( -3.56155281)

So you'll have two real roots:
x = +/- 0.749368276

And two imaginary roots:
x = +/- 1.88720768 i

2006-11-02 08:11:35 · answer #1 · answered by Puzzling 7 · 0 0

( (x^4) + (3(x^2)) - 2 ) = 0
x^4+3x^2-2=0
x^2=(-3+/-sqrt(9+8))/2
x^2=-3/2+/-sqrt17)/2
x^2=.56155

x=+/-.749 2 real roots.
x^2=-3.561
x==/-1.89i 2 imaginary roots.

2006-11-02 16:11:15 · answer #2 · answered by yupchagee 7 · 0 0

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