# To solve biquadratic equation

1. Mar 18, 2013

### Monsterboy

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Mar 18, 2013

### SteamKing

Staff Emeritus
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)

3. Mar 18, 2013

### epenguin

You've done the hard part.

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

4. Mar 19, 2013

### 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 .

Last edited: Mar 19, 2013
5. Mar 19, 2013

### ehild

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

ehild

6. Mar 19, 2013

### Staff: Mentor

7. Mar 19, 2013

### Monsterboy

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?

8. Mar 19, 2013

### 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!

9. Mar 19, 2013

### Monsterboy

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.

10. Mar 19, 2013