You can put this solution on YOUR website!
let y = x^2
your equation of:
3y^2 + 5y - 2 = 0
factor this to get:
(3y-1) * (y+2) = 0
solve for y to get:
y = 1/3
y = -2
since y = x^2, substitute for y to get:
x^2 = 1/3
x^2 = -2
take the square root of both sides of each equation to getr:
x = +/- sqrt(1/3)
x = +/- sqrt (-2)
sqrt (-2) is not a real number so that answer is not valid.
you are left with x = +/- sqrt (1/3).
you need to substitute for x in your original equation to see if these values of x are good.
your original equation is:
substituting + sqrt(1/3) gets:
3*(sqrt(1/3))^4 + 5*(sqrt(1/3))^2 - 2 = 0
this results in:
0 = 0
substituting - sqrt(1/3) gets the same answer because it is being raised to an even power which makes the answer the same as when it was + sqrt(1/3)
the key to solving this problem is recognizing that you have a quadratic formula in disguise.
the quadratic formula is:
ax^2 + bx + c = 0
anytime you see something like ax^4 + bx^2 + c = 0, you can substitute y = x^2 and get the quadratic equation of ay^2 + by + c = 0.
multiples also works.
the equation ax^10 + bx^5 + c = 0 can be solved in the same way.
just make y = x^5 and you get:
ay^2 + by + c = 0
since y = x^5, y^2 = x^5 squared which is equal to x^10
just remember that you have to get back to x at the end after you solved for y.
you can skip the y part if you prefer.
your equation is:
3x^4+5x^2-2=0 which is equivalent to:
3*(x^2)^2 + 5*(x^2)^1 - 2 = 0
you'll get the same answer but setting it equal to y just makes subsequent processing easier on the eyes.
just remember that you have to get back to x for your final answer.