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

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

  • 3
    $\begingroup$ 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. $\endgroup$ Sep 22, 2010 at 23:25
  • $\begingroup$ A typical contest question with this setup asks for the sum of the squares (or cubes) of c and d. $\endgroup$
    – Isaac
    Sep 23, 2010 at 0:11

3 Answers 3


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.


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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    $\begingroup$ "So this Vieta must have been a really smart guy, but you've rediscovered his trick." -LOL, thanks for making my day, Agusti! +1. $\endgroup$ Sep 23, 2010 at 3:59
  • $\begingroup$ My pleasure. :-) $\endgroup$ Sep 23, 2010 at 7:43
  • $\begingroup$ how did you do the fancy algebraic letters? surely it will be y^2 - ay - b = 0 $\endgroup$
    – Jonathan.
    Sep 24, 2010 at 22:02
  • $\begingroup$ 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 = 10 $\endgroup$
    – Jonathan.
    Sep 24, 2010 at 22:45
  • $\begingroup$ @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". $\endgroup$ Sep 25, 2010 at 0:57

HINT $\rm\quad (X - c)\: (X - d) \; =\; X^2 - a\ X + b $


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