*Quadratic Equation...!?

Question

This is a new problem to me, I have never encountered a problem with the "4th power".

12x^4-31x^2+20=0

Plz explain step by step!

Answer 1

let y=x^2
12y^2 - 31y + 20 =0
(3y-4)(4y-5)=0

y = 4/3 or 5/4

x = sqrt(4/3) or sqrt(5/4)

Answer 2

This can be treated as a quadratic equation with x^2 as variable to begin with. Thus,
12x^4-31x^2+20=0
=> 12 (x^2)^2 - 31(x^2) + 20 = 0
=> 12 (x^2)^2 - 16(x^2) - 15(x^2) + 20 = 0
=> 4(x^2) (3x^2 - 4) - 5 (3x^2 - 4) = 0
=> (4x^2 - 5)(3x^2 - 4) = 0
=> 4x^2 = 5 or 3x^2 = 4
=> x = ± √5 / 2 or x = ± 2 / √3.

Answer 3

Divide equation by x^2 giving:
12x^2-31x+20=0
(x-5/4)(x-4/3)=0
x=5/4 or x=4/3

Answer 4

Notice you only have even powers. Replace your x^2 with, say, a y. Then your x^4 is the same as a y^2, right, so then you have:
12y^2-31y+20=0, which is a quadratic equation. Solve it for y as normal. Then, once you have those 2 solutions (if there are 2) notice that those are actually the answers for x^2, so you have to take the positive and negative square roots of those numbers to get your answers for x (there should be 4 answers).

Answer 5

I have no idea if this is right but this is how I would do it!! Good Luck!!

12x^4-31^2+20=0
=-19x^2+20=0
=-19x^2=-20
=19x^2=20
=x^2=20/19
=x= square root of 20/19
=1.0259

Answer 6

replace x² with u. then you have
12u² - 31u + 20 = 0
(3u - 4)(4u - 5) = 0
3u - 4 = 0
u = 4/3 = x²
x = √(4/3) = 2√(1/3) = (2/3)√3 .... or
4u - 5 = 0
u = 5/4 = x²
x = (1/2)√5

Answer 7

I should know this...I'm half way through a maths degree at university.
I will edit if I remember...sorry.