SOLUTION: Please help. Solve the following 15x^4-28x^2+5=0. I know that eventually I have to use the quadratic equation, but I am unsure of how to get the problem into a standard form to use

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help. Solve the following 15x^4-28x^2+5=0. I know that eventually I have to use the quadratic equation, but I am unsure of how to get the problem into a standard form to use      Log On


   

Question 177410: Please help. Solve the following 15x^4-28x^2+5=0. I know that eventually I have to use the quadratic equation, but I am unsure of how to get the problem into a standard form to use. Thank you.
Found 3 solutions by jim_thompson5910, stanbon, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
15x%5E4-28x%5E2%2B5=0 Start with the given equation.


Let z=x%5E2. So this means that z%5E2=%28x%5E2%29%5E2=x%5E4


15z%5E2-28z%2B5=0 Replace x%5E4 with z%5E2 and x%5E2 with "z"


Notice we have a quadratic equation in the form of az%5E2%2Bbz%2Bc where a=15, b=-28, and c=5


Let's use the quadratic formula to solve for z


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%28-28%29+%2B-+sqrt%28+%28-28%29%5E2-4%2815%29%285%29+%29%29%2F%282%2815%29%29 Plug in a=15, b=-28, and c=5


z+=+%2828+%2B-+sqrt%28+%28-28%29%5E2-4%2815%29%285%29+%29%29%2F%282%2815%29%29 Negate -28 to get 28.


z+=+%2828+%2B-+sqrt%28+784-4%2815%29%285%29+%29%29%2F%282%2815%29%29 Square -28 to get 784.


z+=+%2828+%2B-+sqrt%28+784-300+%29%29%2F%282%2815%29%29 Multiply 4%2815%29%285%29 to get 300


z+=+%2828+%2B-+sqrt%28+484+%29%29%2F%282%2815%29%29 Subtract 300 from 784 to get 484


z+=+%2828+%2B-+sqrt%28+484+%29%29%2F%2830%29 Multiply 2 and 15 to get 30.


z+=+%2828+%2B-+22%29%2F%2830%29 Take the square root of 484 to get 22.


z+=+%2828+%2B+22%29%2F%2830%29 or z+=+%2828+-+22%29%2F%2830%29 Break up the expression.


z+=+%2850%29%2F%2830%29 or z+=++%286%29%2F%2830%29 Combine like terms.


z+=+5%2F3 or z+=+1%2F5 Simplify.


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Since we let z=x%5E2, we need to plug that in:


z+=+5%2F3 Start with the first solution for "z"


x%5E2+=+5%2F3 Plug in z=x%5E2


x+=+0%2B-sqrt%285%2F3%29 Take the square root of both sides.


x+=+sqrt%285%2F3%29 or x+=+-sqrt%285%2F3%29 Break up the "plus/minus".


x+=+sqrt%2815%29%2F3 or x+=+-sqrt%2815%29%2F3 Simplify the square root.


So the first two solutions are x+=+sqrt%2815%29%2F3 or x+=+-sqrt%2815%29%2F3


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z+=+1%2F5 Start with the second solution for "z"


x%5E2+=+1%2F5 Plug in z=x%5E2


x+=+0%2B-sqrt%281%2F5%29 Take the square root of both sides.


x+=+sqrt%281%2F5%29 or x+=+-sqrt%281%2F5%29 Break up the "plus/minus".


x+=+sqrt%285%29%2F5 or x+=+-sqrt%285%29%2F5 Simplify the square root.


So the next two solutions are x+=+sqrt%285%29%2F5 or x+=+-sqrt%285%29%2F5


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Answer:


So the four solutions are:

x+=+sqrt%2815%29%2F3, x+=+-sqrt%2815%29%2F3, x+=+sqrt%285%29%2F5 or x+=+-sqrt%285%29%2F5

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
15x^4-28x^2+5=0
---------
15x^4 - 25x^2 - 3x^2 + 5 = 0
5x^2(3x^2 - 5) - (3x^2 - 5) = 0
(3x^2 - 5)(5x^2 - 1) = 0
x^2 = 5/3 or x^2 = 1/5
x = +/-sqrt(5/3) or x = +/-sqrt(1/5)
========================================
Cheers,
Stan H.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
15x%5E4-28x%5E2%2B5=0

The trick is to let t+=+x%5E2. That gives you 15t%5E2+-+28t+%2B+5+=+0 which is a quadratic in standard form.

Look at the discriminant b%5E2-4ac+=+784+-+300+=+484+=+22%5E2. Since the discriminant is a perfect square, the quadratic has rational roots and can be factored:

%285x+-+1%29%283x+-+5%29+=+0

Hence: t+=+1%2F5 or t+=+5%2F3

But remember t+=+x%5E2 so:

x+=+sqrt%281%2F5%29+=+sqrt%285%29%2F5 or x+=+-sqrt%281%2F5%29+=+-sqrt%285%29%2F5

or

x+=+sqrt%285%2F3%29+=+sqrt%2815%29%2F3 or x+=+-sqrt%285%2F3%29+=+-sqrt%2815%29%2F3

Which is 4 roots as you would expect with a quartic.