# Finding two numbers given their sum and their product

Which two numbers when added together yield $16$, and when multiplied together yield $55$.

I know the $x$ and $y$ are $5$ and $11$ but I wanted to see if I could algebraically solve it, and found I couldn't.

In $x+y=16$, I know $x=16/y$ but when I plug it back in I get something like $16/y + y = 16$, then I multiply the left side by $16$ to get $2y=256$ and then ultimately $y=128$. Am I doing something wrong?

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How do you know it's easy if you haven't solved it? Anyway, you're not multiplying by $16$ correctly, and you don't want to multiply by $16$ anyway. Try multiplying by $y$. Also, you get $x = \frac{55}{y}$. – Qiaochu Yuan Jul 16 '12 at 5:10
Note: typically you want to use a matrix to solve a system of linear equations. If you set up the problem as in the answers below and use $xy = 55$, you are no longer using a linear equation. – The Chaz 2.0 Jul 16 '12 at 5:25
The problem is equivalent to solving a monic quadratic equation w/coefficients 16 and 55... – DVD Aug 12 '15 at 21:58

Our two equations are: $$x + y = 16 \tag{1}$$ $$xy = 55\tag{2}$$

Rewriting equation (1) in terms of just $y =$ something, we get:

$$y = 16-x$$

Substituting this into equation (2) leaves us: $$x(16-x) = 55$$ $$16x-x^2=55 \implies x = 5 \ \ \text{or} \ \ 11$$

which can be easily seen by factoring or using the quadratic formula. It follows that $y=11|x=5$ and $y=5|x=11$.

Thus your solutions in terms of $(x,y)$ are $(5,11)$ and $(11,5)$.

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By factoring or using the quadratic formula? Do you have a method of factoring quadratics that doesn't involve finding its roots first? – Henning Makholm Jul 16 '12 at 10:58
@HenningMakholm Simple. You know the factorized quadratic will look like $(x-X)(x-Y)$, and $X$ and $Y$ satisfy... oh crap. – Sean Eberhard Jan 19 '15 at 21:13

We are trying to solve the system of equations $x+y=16$, $xy=55$. Here are a couple of systematic approaches that work in general.

Approach $1$: We will use the identity $(x+y)^2-4xy=(x-y)^2$. In our case, we have $(x+y)^2=256$, $4xy=220$, so $(x-y)^2=36$, giving $x-y=\pm 6$.

Using $x+y=16$, $x-y=6$, we get by adding that $2x=22$, and therefore $x=11$. It follows that $y=5$.

The possibility $x+y=16$, $x-y=-6$ gives nothing new. Adding, we get $2x=10$, so $x=5$, and therefore $y=11$.

Approach $2$: From $x+y=16$, we get $y=16-x$. Substitute for $y$ in $xy=55$. We get $x(16-x)=55$. Simplification gives $x^2-16x+55=0$. The quadratic factors as $(x-5)(x-11)$, so our equation becomes $(x-5)(x-11)=0$, which has the solutions $x=5$ and $x=11$.

But we cannot necessarily rely on there being such a straightforward factorization. So in general after we get to the stage $x^2-16x+55=0$, we would use the Quadratic Formula. We get $$x=\frac{16\pm\sqrt{(-16)^2-4(55)}}{2}.$$
Compute. We get the solutions $x=5$ and $x=11$. The corresponding $y$ are now easy to find from $x+y=16$.

Remarks: $1,$ Recall that the Quadratic Formula says that if $a\ne 0$, then the solutions of the quadratic equation $ax^2+bx+c=0$ are given by $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Your approach was along reasonable lines, but things went wrong in the details. From $xy=55$ we get $x=\frac{55}{y}$. Substituting in the formula $x+y=16$, we get $$\frac{55}{y}+y=16.$$ A reasonable strategy is to multiply through by $y$, getting $55+y^2=16y$, or equivalently $y^2-16y+55=0$. Now we have reached a quadratic equation which is basically the same as the one we reached above.

$2.$ The first approach that we used (presented as an algorithm, and stripped of algebraic notation) goes back to Neo-Babylonian times. The "standard" problem was to find the dimensions of a door, given its perimeter and area.

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Approach 1 is the most elegant answer. (To see how you'd think of using $x-y$, you get to it by looking at the graph of $xy=55$ and $x+y = 16$ plotted on the same axes, and seeing that it's symmetrical about the line $y=x$. So it's natural to change variables and think of the graph as plotted on the axes $y=x$, $y=-x$.) – HTFB Jul 31 '14 at 11:19

Here is another method: suppose you are told that two numbers, $x$ and $y$, have a certain sum $x+y=S$, and a certain product $xy=P$. How to find $S$ and $P$?

We can use the fact that we know how to solve quadratic equations. Notice that $$(t-x)(t-y) = t^2 - (x+y)t + xy = t^2 - St + P.$$

That means that $x$ and $y$ are precisely the solutions to $$t^2 - St + P = 0.$$

In your specific case, $S=16$ and $P=55$. So we want to find the solutions to $$t^2 - 16t + 55 = 0.$$

The quadratic formula gives $$t = \frac{16 \pm\sqrt{256 - 220}}{2} = 8 \pm\frac{1}{2}\sqrt{36} = 8\pm\frac{6}{2} = \left\{\begin{array}{l} 11\\ 5 \end{array}\right.$$ So the two numbers are $5$ and $11$.

(Of course, we often solve quadratic equations $t^2 + at + b=0$ by figuring out by eyeballing two numbers whose product is $b$ and whose sum is $-a$, but we can always use the quadratic formula to take the guessing out of it.)

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The average of x and y is 16/2 = 8, their product is xy = 55 therefore: x, y = 8 plus or minus sqrt of 8 square minus 55 = 8 +/- sqrt of 9 = 8 +/- 3, x, y = 11, 5

by: GeorgeB reference: Vedic Book

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let the sum be S and the product be P

you can get the two numbers x and y using the following formular

x = (S + (S^2 - 4P)^-2)/2

y = (S - (S^2 - 4P)^-2)/2

I've done the math just like above

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should be

x = (S + ((S^2 - 4P)^(1/2)))/2

y = (S - ((S^2 - 4P)^(1/2)))/2

i.e. raising to the power of a half is taking the square root. raising to the power of -2 is the reciprocal of the square.

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– Shaun Jul 31 '14 at 9:57

$$x + y = 16 ... (1*)$$

$$x * y = 55 ... (2*)$$

Use identity: $$( x-y)^2 = = ( x+y)^2 - 4 *x*y = ... (3*)$$

$$x - y = 256 - 4 * 55 = \sqrt{36} \rightarrow x -y = \pm \, 6 ...(4*)$$

Find half sum and half difference of equations (1*) and (2*):

$$(x,y) = (11,5) , \, (5,11) .. (*5)$$

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My favorite way

Square the sum to get $S^2=a^2+2ab+b^2$.

Get the squared difference $D^2=a^2-2ab+b^2$ by deducing four times the product.

Then $a,b=\dfrac{S\pm D}2$.

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Mentally, $D^2=256-220=6^2$, then $a,b=5,11$. – Yves Daoust May 7 '15 at 20:03

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