# a question

• Oct 28th 2006, 01:46 PM
fima2001
a question
for what values of the parameter m there will be
4 answers in wich can be arranged so the subtraction of each 2 answers
will be permanent
• Oct 28th 2006, 02:58 PM
CaptainBlack
Quote:

Originally Posted by fima2001
for what values of the parameter m there will be
4 answers in wich can be arranged so the subtraction of each 2 answers
will be permanent

Maybe someone else will be able to tell us what this means, but to me
it seems incomprehensible. Can you clarify what it is you are asking?

Thanks

RonL
• Oct 28th 2006, 04:35 PM
topsquark
Quote:

Originally Posted by fima2001
for what values of the parameter m there will be
4 answers in wich can be arranged so the subtraction of each 2 answers
will be permanent

To start you off, this is a biquadratic equation. So:

$x^4 + 6mx^2 + (3m+8)^2 = 0$

You can solve this for $x^2$ using the quadratic formula:

$x^2 = \frac{-6m \pm \sqrt{36m^2 - 4(3m+8)^2}}{2}$

$x^2 = \frac{-6m \pm \sqrt{36m^2 - 36m^2 - 192m - 256}}{2}$

$x^2 = \frac{-6m \pm \sqrt{-192m - 256}}{2}$

$x^2 = \frac{-6m \pm 16 \sqrt{-3m - 4}}{2}$

$x^2 = -3m \pm 8 \sqrt{-3m - 4}$

Now take the square root of both sides to get your 4 solutions:
$x = \pm \sqrt{ -3m \pm 8 \sqrt{-3m - 4}}$ (Where the two $\pm$ symbols are independant of each other.)

To continue we need to know what you mean by "the subtraction of each 2 answers will be permanent."

-Dan
• Oct 28th 2006, 08:16 PM
Soroban
Hello, all!

I think I have an interpretation . . . and a solution.

Quote:

For what values of the parameter $m$ will the equation
have four roots in arithmetic progression?

. . $x^4 + 6mx^2 + (3m+8)^2\;=\;0$

Let the four roots be: . $a,\:a+d,\:a+2d,\:a+3d$

The sum of the roots is equal to the negative of the coefficient of $x^3$.
. . Hence: . $a + (a+d)+(a+2d)+(a+3d) \:=\:0$
. . And we have: . $4a + 6d\:=\:0\quad\Rightarrow\quad d = -\frac{2}{3}a$ [1]

The product of the roots is equal to the constant term.
. . Hence: . $a(a+d)(a+2d)(a+3d) \:=\:(3m+8)^2$

Substitute [1]: . $a\left(a - \frac{2}{3}a\right)\left(a - \frac{2}{3}a\right)\left(a - 2a\right)\;=\;(3m+8)^2$

. . $a\left(\frac{a}{3}\right)\left(-\frac{a}{3}\right)\left(-a\right) \:=\:(3m+8)^2\quad\Rightarrow\quad a^4 = 9(3m+8)^2\quad\Rightarrow\quad a^2 = 3(3m+8)^2$ [2]

The sum of the the roots, taken in pairs, is the coefficient of $x^2$.
$a(a+d)+a(a+2d)+a(a+3d)+(a+d)(a+2d)+$ $(a+d)(a+3d)+(a+2d)(a+3d)\:=\:6m$

Substitute [1]: . $a\left(\frac{a}{3}\right) + a\left(-\frac{a}{3}\right) + a(-a) + \left(\frac{a}{3}\right)\left(-\frac{a}{3}\right) + \left(\frac{a}{3}\right)\left(-a\right) + \left(-\frac{a}{3}\right)\left(-a\right) \:=\:6m$

and we have: . $-\frac{10}{9}a^2\:=\:6m\quad\Rightarrow\quad a^2 \:=\:-\frac{27}{5}m$ [3]

Equate [2] and [3]: . $3(3m + 8) \:=\:-\frac{27}{5}m\quad\Rightarrow\quad \boxed{m = -\frac{5}{3}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

From [3]: . $a^2\:=\:-\frac{27}{5}\left(-\frac{5}{3}\right)\:=\;9\quad\Rightarrow\quad a = \pm3$

And from [1]: . $d \:=\:-\frac{2}{3}(\pm3)\:=\:\mp2$

Therefore, the roots are: . $-3,\:-1,\:1,\:3$

• Oct 28th 2006, 10:58 PM
fima2001
i ment
there will be 4 answers and the subtraction of each 2 close
roots will be the same resolt