3x^4+5x^2-2=0? HELP!!!!!!!!!!?

Question

solve using the quadratic equation!

Answer 1

3x^4 + 5x² - 2 = 0
Let y = x²
3y² + 5y - 2 = 0
3y² + 6y - y - 2 = 0
3y(y+2)-1(y+2) = 0
(y+2)(3y-1) = 0
y = {-2, 1/3}

x = sqrt(y)
x = {±(√2)i, ±1/√3}

Using the quadratic formula:
ax² + bx + c = 0
x = (-b±√(b²-4ac))2a

3(x²)² + 5x² - 2 = 0
a = 3
b = 5
c = -2

x² = (-5±√(5²+24))/6
= (-5±√49)/6
= (-5±7)/6
= {2/6, -12/6}
= {1/3, -2}

x = {±√(1/3), ±√(-2)}
x = {±(√2)i, ±1/√3}

Answer 2

Since you have no odd powers of x, try replacing k = x²

3k² + 5k - 2 = 0

Now you can factor this easily as:
(3k - 1)(k + 2) = 0

So k = 1/3 or k = -2

Substitute x² back in and you have:
x² = 1/3 or x² = -2

Taking the square root of each of these you get:
x = ±√(1/3) = ±(√3) / 3
x = ±√(-2) = ±(√2)i

You have 4 roots (2 real and 2 imaginary)

Real roots:
√3 / 3
-√3 / 3

Imaginary roots:
√2 i
-√2 i

Answer 3

Use the Quadratic Formula: x equals negative b, plus or minis square root b squared minus 4ab all over 2a. First, subtract 2 from both sides. now u have 3x^4 + 5x^2=2. the first term is a, the second term is b and the third term is c. substitute all a, b and c with the quadratic formula. that should give you the answer. = )

Answer 4

Instead of x^2, replace it with t = x^2.

You then look at the equation:
3t^2 + 5t - 2 = 0

Solutions:
t = (-5 + sqrt(5*5 - 4*3*(-2)) / (3*2) = (-5 + 7) / 6 = 2 / 6 = 1 / 3 = 0.333...
t = (-5 - sqrt(5*5 - 4*3*(-2)) / (3*2) = (-5 - 7) / 6 = -12 / 6 = -2

Since t = x^2:
x = sqrt(-2) = 2*i or -2*i
or
x = sqrt(1/3).

Hope this helps. :)

* sqrt = square root *

Answer 5

PEMDAS! Please Excuse My Dear Aunt Sally! 0
2-2 = 0 so 4=5x^0 = 0 so 3x^0 = 0. Its been a long time. don't laugh at me if I'm wrong. lol!

Answer 6

what are you trying to solve for and what are you using. I'm pretty good at this, so just let me know

Answer 7

what the heck are you trying to do?
find the roots, solve for x? i have no idea.

Answer 8

answer is 20

Answer 9

??? >123<323....444 :-)

Answer 10

thats pretty scary