Help with Math, test tomorrow.?

Question

Solve each equation.
1) 3x^4 - 3x^2 -6 = 0

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Answer 1

First factor out the greatest common factor. It is 3
3(x^4 - x^2 - 2) = 0
What factors of -2 add to -1? The factors are -2 and +1
3(x^2 +1)(x^2 -2) = 0
Set each part equal to zero.
x^2 +1 = 0 or x^2 - 2 = 0
x^2 + 1 has no real solutions...just imaginary. They are i and -i.
x^2 -2 = 0 when x = sqrt(2) or -sqrt(2)

Answer 2

3x^4 - 3x^2 -6
= 3(x^4 - x^2 - 2) = 0
3(x^2 +1)(x^2 -2) = 0
so
x^2 +1 = 0 or x^2 - 2 = 0
x^2 + 1 >0 so there is no solution
x^2 -2 = 0, then x = sqrt(2) or -sqrt(2)


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Answer 3

3x^4 -3X2 -6 =0

Put y = x^2 (*)

We have : 3y^2 -3y -6= 0 ( /3)

y^2 -y -2 = 0

( y-2) (y+1) (Factorization)

y-2=0 => y=2
y+1=0=> y= -1

So we substitute y=2 and y= -1 in (*)

and we obtain

x^2 = 2 => x= +/- sqrt 2

x^2 = -1 => x= +/- sqrt(-1) = +/- i (because
i^2= -1)

Final Answer: { +/-sqrt 2, +/-i }

Verification:

x= 3(+sqrt2)^4 -3(sqrt2)^2 -6=0 => 3 x 4 -6-6=0
(true)

x= 3(-sqrt2)^4 -3(-sqrt2)^2-6=0 =>
3 x 4-6-6=0 (true)

x=i => 3(i)^4 -3(i)^2 -6 = 0 (n.b.: i^2 = -1)
3(1) -3(-1) -6 =0 (true)

x= -i => 3(-i)^4 -3(-i)^2 -6 =0 (n.b.:(-i)^2= -i x-i
= i^2 = -1)
3(1) -3(-1) -6 =0 (true).

Answer 4

3x^4 - 3x^2 -6 = 0
x^4-x^2-2=0
(x^2-2)(x^2+1)=0
x=sqrt(2)
x=-sqrt(2)
x=i
x=-i are the 4 roots