Maths find roots?

Chewbacca.

please someone tell me the easiest way to find roots of
8x^4 - 7x^2 + 1 =0
?

3DataModem

Set y = x^2

your equation is 8y^2 - 7y + 1 = 0

solve for y using one of the normal methods

then derive x (by simply finding the squared root of both instances of y)

be careful - there may be up to 4 distinct solutions

there are

jumpguy

To find the first roots, you need to go trial and error. Try substituting 1 or -1 for x, if you don't get zero, try substituting 2 or -2 for x, and so on. These problems rarely go beyond 3, so that shouldn't take too long.

Once you've found what goes in for x, lets say 2 goes in for x, then x=2. Therefore your factor would be x-2.

Divide the equation you started with by your factor (say if it's x-2), and you'll find the other roots.

Good luck!

EDIT: Just saw 3DataModem's post - am I wrong?

3DataModem

So that's...

y = ( 7 +\- sqr ( 17 ) ) / 16 )

x = ( +\- sqr ( 7 +\- sqr ( 17 ) ) / 16 )

...ish

3DataModem

Jumpguy - for a lot of problems your guesswork/interpolation method works fine but will prob give only one or two solutions.

Note that one or two solutions are likely to be some multiple of "i".... tricky to find by hand.

Chewbacca.

i'm a bit lost as I can only get x=1 or x=0.35?

LeixlipRed

jumpguy wrote: »

To find the first roots, you need to go trial and error. Try substituting 1 or -1 for x, if you don't get zero, try substituting 2 or -2 for x, and so on. These problems rarely go beyond 3, so that shouldn't take too long.

Once you've found what goes in for x, lets say 2 goes in for x, then x=2. Therefore your factor would be x-2.

Divide the equation you started with by your factor (say if it's x-2), and you'll find the other roots.

Good luck!

EDIT: Just saw 3DataModem's post - am I wrong?

Dangerous assumptions to be making there!! Only works if the roots are integers and if they're close to zero. For a quadratic of this form (missing the X^3 and x term) you use a substitution, say y=x^2 to solve.