Answer:
The numbers are
[tex]4+2\sqrt{6}[/tex] and [tex]-4+2\sqrt{6}[/tex]
Step-by-step explanation:
Let
x ------> one number
(x-8) ------> the another number
we know that
[tex]x^{2} +(x-8)^{2}=80[/tex]
[tex]x^{2} +x^{2}-16x+64=80[/tex]
[tex]2x^{2}-16x+64-80=0[/tex]
[tex]2x^{2}-16x-16=0[/tex]
Simplify
[tex]x^{2}-8x-8=0[/tex]
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}-8x-8=0[/tex]
so
[tex]a=1\\b=-8\\c=-8[/tex]
substitute in the formula
[tex]x=\frac{8(+/-)\sqrt{-8^{2}-4(1)(-8)}} {2(1)}[/tex]
[tex]x=\frac{8(+/-)\sqrt{96}} {2}[/tex]
[tex]x=\frac{8(+/-)4\sqrt{6}} {2}[/tex]
[tex]x=4(+/-)2\sqrt{6}[/tex]
[tex]x1=4(+)2\sqrt{6}[/tex]
[tex]x2=4(-)2\sqrt{6}[/tex]
so
the solution must be a positive real number
[tex]x=4(+)2\sqrt{6}[/tex]
[tex]x-8=4(+)2\sqrt{6}-8[/tex] ------> [tex]x-8=-4(+)2\sqrt{6}[/tex]
