SOLUTION: Help me to solve: 10x^4 - 19x^2 + 6 = 0

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Question 423723: Help me to solve:
10x^4 - 19x^2 + 6 = 0
Found 2 solutions by stanbon, dnanos:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
10x^4 - 19x^2 + 6 = 0
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10x^4-15x^2-4x^2+6 = 0
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5x^2(2x^2-3)-2(2x^2-3) = 0
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(2x^2-3)(5x^2-2) = 0
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Solve:
2x^2-3 = 0
x^2 = 3/2
x = +-sqrt(6)/2
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Solve:
5x^2-2 = 0
x^2 = 2/5
x = +-sqrt(10)/5
==============================
Cheers,
Stan H.
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Answer by dnanos(83) About Me  (Show Source):
You can put this solution on YOUR website!
let x^2=y We have to find y and then x...
The general formula for ay%5E2%2Bby%2Bc=0 is y=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
Where a=10,b=-19, and c=6
y=%28-%28-19%29%2B-sqrt%28%28-19%29%5E2-4%2A10%2A6%29%29%2F%282%2A10%29
y=%2819%2B-sqrt%28361-240%29%29%2F20=
%2819%2B-sqrt%28121%29%29%2F20=
%2819%2B-sqrt%2811%5E2%29%29%2F20=
We obtain two values for y:
y=3/2 or y=8/20=2/5=0.4
Each one of them gives two values for x:
x=0%2B-sqrt%28y%29
x=sqrt%283%2F2%29%29%29
or
x=-sqrt%283%2F2%29%29%29
or
x=sqrt%280.4%29
or
x=-sqrt%280.4%29
So we have four solutions:
sqrt%283%2F2%29%29%29;-sqrt%283%2F2%29%29%29;sqrt%280.4%29;-sqrt%280.4%29