# Algebra

3x^4-31x^2+28=0 Solve the equation??

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1. you will need to use the quadratic equation

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posted by Do
2. let y = x^2
then
3 y^2 -31 y + 28 = 0
(3y-28)((y-1) = 0
y = 1 or 28/3
so
x^2 = 1
x = 1 or -1
x^2 = 28/3
x = 2 sqrt(7/3) or -2sqrt(7/3)

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posted by Damon
3. let x^2 = a
so we have
3a^2 - 31a + 28 = 0
(3a-28)(a-1) = 0
a = 28/3 or a = 1
so
x^2 = 28/3
x = ± 2√7/√3 or a = ±1
or
x = ± 2√21/3 , a = ±1

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posted by Reiny

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