Answer:
The numbers are
[tex]6+2\sqrt{15}i[/tex] and [tex]6-2\sqrt{15}i[/tex]
Step-by-step explanation:
Let
x and y -----> the numbers
we know that
[tex]x+y=12[/tex] -----> [tex]y=12-x[/tex] ------> equation A
[tex]xy=96[/tex] ----> equation B
substitute equation A in equation B and solve for x
[tex]x(12-x)=96\\12x-x^{2}=96\\x^{2} -12x+96=0[/tex]
Solve the quadratic equation
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} -12x+96=0[/tex]
so
[tex]a=1\\b=-12\\c=96[/tex]
substitute
[tex]x=\frac{-(-12)(+/-)\sqrt{-12^{2}-4(1)(96)}} {2(1)}[/tex]
[tex]x=\frac{12(+/-)\sqrt{-240}} {2}[/tex]
Remember that
[tex]i^{2}=\sqrt{-1}[/tex]
[tex]x=\frac{12(+/-)\sqrt{240}i} {2}[/tex]
[tex]x=\frac{12(+/-)4\sqrt{15}i} {2}[/tex]
Simplify
[tex]x=6(+/-)2\sqrt{15}i[/tex]
[tex]x1=6+2\sqrt{15}i[/tex]
[tex]x2=6-2\sqrt{15}i[/tex]
we have two solutions
Find the value of y for the first solution
For [tex]x1=6+2\sqrt{15}i[/tex]
[tex]y=12-x[/tex]
substitute
[tex]y1=12-(6+2\sqrt{15}i)[/tex]
[tex]y1=6-2\sqrt{15}i[/tex]
Find the value of y for the second solution
For [tex]x2=6-2\sqrt{15}i[/tex]
[tex]y2=12-x[/tex]
substitute
[tex]y2=12-(6-2\sqrt{15}i)[/tex]
[tex]y2=6+2\sqrt{15}i[/tex]
therefore
The numbers are
[tex]6+2\sqrt{15}i[/tex] and [tex]6-2\sqrt{15}i[/tex]
