SOLUTION: I have a hard exponent quadratic question. 5x^2/3 + 2x^4/3 -13 =0 My first inclination was to rearrange the terms to... 2x^4/3 + 5x^2/3 -13 =0 This puts it in the right

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Question 353894: I have a hard exponent quadratic question.
5x^2/3 + 2x^4/3 -13 =0
My first inclination was to rearrange the terms to...
2x^4/3 + 5x^2/3 -13 =0
This puts it in the right order for substitution / y= x^2/3
Thus... 2y^2 + 5y -13 =0
This is where I'm stuck. I don't know how to unfold the answer with the unusual powers.
Thanks for your help.
Neil
Found 3 solutions by Earlsdon, Alan3354, ewatrrr:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you are ok so far but now you can use the quadratic formula (y+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a) to solve for y. In this problem, a = 2, b = 5, and c = -13, so...
y+=+%28-5%2B-sqrt%285%5E2-4%282%29%28-13%29%29%29%2F2%282%29
When you evaluate this you will get:
y+=+-4.089 or 1.589
Now you will need to substitute y+=+x%5E%282%2F3%29, so...
x%5E%282%2F3%29+=+-4.089 or x%5E%282%2F3%29+=+1.589
To find x, raise both sides of each solution to the 3%2F2 (the reciprocal of 2%2F3%29) power.
x+=+%28-4.089%29%5E%283%2F2%29 or x+=+%281.589%29%5E%283%2F2%29 Use your calculator to evaluate to get:
x+=+-8.268i or x+=+2
It would be advisable to check these solutions in your original equations.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
5x^2/3 + 2x^4/3 -13 =0
My first inclination was to rearrange the terms to...
2x^4/3 + 5x^2/3 -13 =0
This puts it in the right order for substitution / y= x^2/3
Thus... 2y^2 + 5y -13 =0
This is where I'm stuck. I don't know how to unfold the answer with the unusual powers.
Thanks for your help.
-------------------------
Solve your equation for y:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B5x%2B-13+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%285%29%5E2-4%2A2%2A-13=129.

Discriminant d=129 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+129+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+129+%29%29%2F2%5C2+=+1.58945417290014
x%5B2%5D+=+%28-%285%29-sqrt%28+129+%29%29%2F2%5C2+=+-4.08945417290014

Quadratic expression 2x%5E2%2B5x%2B-13 can be factored:
2x%5E2%2B5x%2B-13+=+%28x-1.58945417290014%29%2A%28x--4.08945417290014%29
Again, the answer is: 1.58945417290014, -4.08945417290014. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B5%2Ax%2B-13+%29

y = -5/4 ± sqrt(129)/4
x^(2/3) = -5/4 ± sqrt(129)/4
It's messy, but it's just arithmetic from here.
Cube both sides:
x^(2/3) = -5/4 + sqrt(129)/4
x^2 = (1/16)*(-515 + 51sqrt(129))
x+=+%2B+%281%2F4%29%2Asqrt%28-515+%2B+51sqrt%28129%29%29
x+=+-+%281%2F4%29%2Asqrt%28-515+%2B+51sqrt%28129%29%29
------------------
Can you do rest?

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
HI,
2x^4/3 + 5x^2/3 -13 =0
.
2x^4 + 5x^2 -39 =0
*Note: This is a bi-quadratic form of an equation. In General:
The bi-quadratic equation is ax4 + bx2 + c = 0 and as it can be writen as:.
.
a%28x2%29%5E2%2Bbx2%2Bc=0 has the roots:
.
x%5E2+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
x%5E2+=+%28-5+%2B-+sqrt%28+5%5E2-4%2A2%2A%28-39%29+%29%29%2F%282%2A2%29+
.
x%5E2+=+%28-5+%2B-+sqrt%28+5%5E2-4%2A2%2A%28-39%29+%29%29%2F%282%2A2%29+
.
x%5E2+=+%28-5+%2B-+sqrt%28+337%29%29%2F4+
.
x+=+sqrt%28%28-5+%2B-+sqrt%28+337%29%29%2F4+%29
.
x+=+%28-5+%2B+sqrt%28+337%29%29%2F4+ or x+=+%28-5+-+sqrt%28+337%29%29%2F4+%29
or
x+=+-%28%28-5++%2B+sqrt%28+337%29%29%2F4%29+ or x+=+-%28%28-5++-+sqrt%28+337%29%29%2F4%29+