Question

rectFind two positive numbers whose difference is 11 and whose product is 432.

Answer 1

Hello!

Let's write these sentences as a system, considering the numbers as x and y:

• difference is 11

(x - y) = 11

• product is 432

(x * y) = 432

System:

$\begin{cases}x-y=11{} \\ x\times y={\text{ }432}\end{cases}$

First, let's isolate the x variable in the first equation:

$x=11+y$

Now, let's replace where's x in the second equation with this obtained expression:

$\begin{gathered} x\times y=432 \\ (11+y)\times y=432 \end{gathered}$

Let's solve using the distributive property:

$\begin{gathered} (11+y)\times y=432 \\ 11y+y^2=432 \\ y^2+11y-432=0 \end{gathered}$

Let's find the coefficients a, b and c:

• a = 1

,

• b = 11

,

• c = - 432

Now, let's solve this quadratic equation that we obtained, using the Quadratic Formula:

$\begin{gathered} \Delta=b^2-4\times a\times c \\ \Delta=11^2-4\times1\times(-432) \\ \Delta=121+1728 \\ \Delta=1849 \end{gathered}$$y=\frac{-b\pm\sqrt{\Delta}}{2\times a}=\frac{-11\pm\sqrt{1849}}{2\times1}=\frac{-11\pm43}{2}$

In this step, as we have + - in the formula we must divide it into two possible values:

$\begin{gathered} y^{\prime}=\frac{-11+43}{2}=\frac{32}{2}=16 \\ \\ y“^=\frac{-11-43}{2}=\frac{-54}{2}=-27\text{ \lparen negative, so disregard this value\rparen} \end{gathered}$

As the exercise asks for positive numbers, we found y = 16.

Now, we have to find X too. To obtain it, we just have to replace y with 16 in any of the equations. I'll do it in the first equation, look:

$\begin{gathered} x-y=11 \\ x-16=11 \\ x=11+16 \\ x=27 \end{gathered}$

So, the answer will be:

(x, y) = (27, 16).