Question

Find two integers whose sum is 16 and product is 55

Answer 1

Since the sum of the two numbers is 16, let one among them be $\sf{x}$, so that the other will be $\sf{16-x.}$

Given that their product is 55. Then,

$\longrightarrow\sf{x(16-x)=55}$

$\longrightarrow\sf{16x-x^2=55}$

$\longrightarrow\sf{x^2-16x+55=0}$

$\longrightarrow\sf{x^2-11x-5x+55=0}$

$\longrightarrow\sf{x(x-11)-5(x-11)=0}$

$\longrightarrow\sf{(x-11)(x-5)=0}$

$\Longrightarrow\sf{x=11\quad OR\quad x=5}$

If $\sf{x=11,}$ then the other number will be $\sf{16-11=5.}$

If $\sf{x=5,}$ then the other number will be $\sf{16-5=11.}$

However, $\bf{11}$ and $\bf{5}$ are the numbers.

Answer 2

Let the numbers be x and y.

Given

Sum of 2 numbers = 16

$\implies x + y = 16$

Product of 2 numbers = 55

$\implies xy = 55$

Solving, we get,

$x = 16-y\\\implies y(16-y) = 55\\\implies -y^{2} +16y = 55\\\implies -y^{2} +16y - 55 = 0\\\implies y^{2} - 16y + 55 = 0\\\implies y^{2} -5y-11y+55=0\\\implies y(y-5)-11(y-5)=0\\\implies (y-11)(y-5)=0$

If y = 11, x = 5

If x = 5, y = 11