How Do You Solve a Biquadratic Equation with Mixed Roots?

[Monsterboy]

Homework Statement

144x^4 -40x^2 -639=0

Homework Equations

there are no clues about the roots ( in A.P,G.P or whatever)

The Attempt at a Solution

there are 2 real roots and 2 complex , i got the complex roots by solving it as a quadratic equation in x^2

144((x^2)^2) -40x^2 -639=0

i don't know how to get the remaining real roots.

 

[SteamKing]

If you have solved for the two complex roots, then you should at least be able to form a quadratic factor using those roots. Divide the original fourth degree polynomial by the quadratic factor (Hint: there should be no remainder)

 

[epenguin]

You've done the hard part.

You've found out values of x². And you ask what values of x they correspond to?

 

[Monsterboy]

i was not clear

by looking at the equation ,we can know that


the term \sqrt(b^2 -4ac) is a square root of negative number

so x^2 itself is complex ,so x^2 has 2 complex roots(conjugate pairs) so the values of x are +/- roots of these two complex roots which leads to 4 complex roots but in my book two real roots are being given + or - 3/2 and it's right, i don't know how to get this .

 

[ehild]

i was not clear

by looking at the equation ,we can know that


the term \sqrt(b^2 -4ac) is a square root of negative number

No, it is not. a=144, c=-639. -4ac>0

ehild

 

[NascentOxygen]

there are no clues about the roots ( in A.P,G.P or whatever)

Things are often clearer when you can see where you are headed.
http://m.wolframalpha.com/input/?i=solve++144x^4+-+40x^2+-639+%3D+0&x=0&y=0&js=0[/color]

 

[Monsterboy]

If you have solved for the two complex roots, then you should at least be able to form a quadratic factor using those roots. Divide the original fourth degree polynomial by the quadratic factor (Hint: there should be no remainder)

i took the two complex values(conjugate pairs) of x^2 and multiplied them to get x^4(real number) and i found the fourth root of that number and i got 1.45 ,i should have got 1.5 ,do you think ,if i had taken more number of digits after decimal places in the calculation ,i would have got 1.5?

 

[epenguin]

You seem to be looking at relatively complicated things and forgetting simple things that you know.

What results have you got for x² ?

Since you are given the answer that tells you that one of these values should be 9/4. If you have not got that you have probably made mistake somewhere.

Your equation you can easily see has one positive and one negative real root for x².

That is given by Descartes' rule, or more simply the LHS is negative at x² = 0 and positive at high positive or negative x², so that is the only possibility.

Then the negative x² gives imaginary x's, the positive one real x's!

 

[Monsterboy]

No, it is not. a=144, c=-639. -4ac>0

ehild

oh great ,what a bloody mistake! , i copied the question wrong in my notebook and wasted a lot time sorry, i had put +639 in the equation.

 

[Monsterboy]

Things are often clearer when you can see where you are headed.
http://m.wolframalpha.com/input/?i=solve++144x^4+-+40x^2+-639+%3D+0&x=0&y=0&js=0[/color]

Nice link!

at least, i solved the wrong question correctly ,i got 4 complex roots
http://m.wolframalpha.com/input/?i=solve++144x^4+-+40x^2++639+=+0&x=0&y=0