Question

The sum of two numbers is 12, their product is 96. Compute these two numbers. Explain.​

Answer 1

Answer:

The numbers are

$6+2\sqrt{15}i$   and  $6-2\sqrt{15}i$

Step-by-step explanation:

Let

x and y -----> the numbers

we know that

$x+y=12$ -----> $y=12-x$ ------> equation A

$xy=96$ ----> equation B

substitute equation A in equation B and solve for x

$x(12-x)=96\\12x-x^{2}=96\\x^{2} -12x+96=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} -12x+96=0$  

so

$a=1\\b=-12\\c=96$

substitute

$x=\frac{-(-12)(+/-)\sqrt{-12^{2}-4(1)(96)}} {2(1)}$

$x=\frac{12(+/-)\sqrt{-240}} {2}$

Remember that

$i^{2}=\sqrt{-1}$

$x=\frac{12(+/-)\sqrt{240}i} {2}$

$x=\frac{12(+/-)4\sqrt{15}i} {2}$

Simplify

$x=6(+/-)2\sqrt{15}i$

$x1=6+2\sqrt{15}i$

$x2=6-2\sqrt{15}i$

we have two solutions

Find the value of y for the first solution

For $x1=6+2\sqrt{15}i$

$y=12-x$

substitute

$y1=12-(6+2\sqrt{15}i)$

$y1=6-2\sqrt{15}i$

Find the value of y for the second solution

For $x2=6-2\sqrt{15}i$

$y2=12-x$

substitute

$y2=12-(6-2\sqrt{15}i)$

$y2=6+2\sqrt{15}i$

therefore

The numbers are

$6+2\sqrt{15}i$   and  $6-2\sqrt{15}i$