Numbers problems

1. Jun 30, 2010

DrummingAtom

It's from Cardano's Algebra book.

Divide 10 into 2 parts whose product is 40.

Is this just called Algebra? I've never seen problems like this before. How would you begin solving it? Thanks.

2. Jun 30, 2010

Pere Callahan

Call the two parts x and y. Then in my understanding the question asks to determine x and y such that x+y=10 and xy=40. Of course there is no reason why such numbers should exist.

3. Jun 30, 2010

Bill Simpson

Hint: In the Reals there may be no reason, but in the Complexes you may find something.

4. Jun 30, 2010

Dickfore

If there are such numbers, then they can be written as $x$ and $10 - x$. Their product is 40, so we have the equation:

$$x (10 - x) = 40$$

This is a quadratic equation. It reduces to the normal form:

$$x^{2} - 10 x + 40 = 0$$

Completing the square of the quadratic trinomial, we get:

$$(x - 5)^{2} + 15$$

This cannot be less than 15, so there are no such real numbers.

5. Jul 1, 2010

Staff: Mentor

Which is not to say that there aren't solutions to the equation. They just don't happen to be real.

6. Jul 1, 2010

Dickfore

Which means his original problem has no solution. He wasn't looking for any numbers, he was looking for 10 to be divided into two parts.

7. Jul 1, 2010

Pere Callahan

And these parts are not a priori restricted to be positive.

8. Jul 1, 2010

Staff: Mentor

Or even real.

According the the Wikipedia article (http://en.wikipedia.org/wiki/Gerolamo_Cardano) I glanced at yesterday, Cardano "acknowledged the existence of what are now called imaginary numbers, although he did not understand their properties".

9. Jul 1, 2010

Pere Callahan

That's what I meant to write.
Then the question is, if he didn't understand their properties, did he know how to multiply complex numbers? He could certainly, without any knowledge write down an expression for x and y involving roots of negative numbers. And complex numbers written in the form of roots of negative numbers are easily multiplied so he could easily check that his result was correct.

10. Jul 1, 2010

CRGreathouse

He understood how to multiply them and add them, just denied that they were meaningful.

11. Jul 2, 2010

Dickfore

Please show where the OP allowed for this possibility?