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I have $2$ numbers $a, b$. I need a formula (or a how to) to find which $2$ numbers $c,d$ will add together to give a and times together to give $b$. So

$c + d = a$
$c \cdot d = b$

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    You will end up with a quadratic equation when you replace either of c or d in the first equation with something derived from the second equation.2010-09-22
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    A typical contest question with this setup asks for the sum of the squares (or cubes) of c and d.2010-09-23

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Consider the quadratic

$$(x-c)(x-d) = x^2 -(c+d)x + cd = x^2 - ax +b$$

Thus the roots of $$x^2 -ax + b = 0$$ are your numbers $c$ and $d$.

The roots are given by

$$\frac{ a \pm \sqrt{a^2 -4b}}{2}$$

Check this page out for more information: Quadratic Equation.

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    Short version: it's the [Vieta formulae](http://mathworld.wolfram.com/VietasFormulas.html) for the quadratic case.2010-09-22
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Since I'm not so smart to imagine that the solution has anything to do with Vieta formulae or roots of polynomials, ;-) I'd try the following. First, let me call your unknowns $c,d$ with letters that look like unkowns (I've tried to do the problem with $c,d$ and at a certain point I forgot which among $a,b,c,d$ were the unkowns and which the data). So let $x = c$ and $y =d$. Your system becomes

$$ \begin{align} x + y &= a \\ xy &= b \ . \end{align} $$

Now, do what you would do in those cases. For instance, solve the first equation for $x$ in terms of $y$:

$$ x = a -y \ . $$

Then substitute this expression for $x$ into the second equation:

$$ (a-y)y = b \ . $$

And, surprise!, you've got a polynomial equation

$$ y^2 -ay + b = 0 $$

that looks like the ones I've been told. (So this Vieta must have been a really smart guy, but you've rediscovered his trick.)

And, of course, once you've got the solutions $y$ of this second degree equation, you should use $x = a-y$ to find $x$... (Just to find out that if $y$ is one of the roots of the last polynomial equation, $x$ is the other one.)

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    "So this Vieta must have been a really smart guy, but you've rediscovered his trick." -LOL, thanks for making my day, Agusti! +1.2010-09-23
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    My pleasure. :-)2010-09-23
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    how did you do the fancy algebraic letters? surely it will be y^2 - ay - b = 02010-09-24
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    sorry I've tried what you said as best as I could but I don't reall understand, could you possibly do an example where a = 7 and b = 102010-09-24
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    @Jonathan. Well, in this case you'll obtain x= 5 and y=2: 5+2= 7 and 5 x 2 = 10. As for the letters, I'm using LateX: click on the link beside "edited" and then on the one that says "view source".2010-09-25
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    Yes I know it's 5 and 2, that's why I asked for such simply numbers, but how I use your answer to get the 5 and 2? Thanks2010-09-26
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    Solve $x+y= 7$ for $x$ in terms of $y$: $x = 7-y$. Substitute this expression for $x$ in $xy = 10$. You get $(7-y)y = 10$. Which is the same as the second degree polynomial equation $y^2 -7y + 10 = 0$. Find the roots of this last equation using the formula Moron reminds you in his answer.2010-09-26
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HINT $\rm\quad (X - c)\: (X - d) \; =\; X^2 - a\ X + b $