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- October 28th 2006, 01:46 PM #1

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- October 28th 2006, 02:58 PM #2

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- October 28th 2006, 04:35 PM #3
To start you off, this is a biquadratic equation. So:

You can solve this for using the quadratic formula:

Now take the square root of both sides to get your 4 solutions:

(Where the two 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

- October 28th 2006, 08:16 PM #4

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Hello, all!

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

How about this?

For what values of the parameter will the equation

have four roots in arithmetic progression?

. .

Let the four roots be: .

The sum of the roots is equal to the negative of the coefficient of .

. . Hence: .

. . And we have: .**[1]**

The product of the roots is equal to the constant term.

. . Hence: .

Substitute**[1]**: .

. .**[2]**

The sum of the the roots, taken in pairs, is the coefficient of .

Substitute**[1]**: .

and we have: .**[3]**

Equate**[2]**and**[3]**: .

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

Check

From**[3]**: .

And from**[1]**: .

Therefore, the roots are: .

- October 28th 2006, 10:58 PM #5

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