Answer :
Answer:
The numbers are
[tex]-2+3\sqrt{5}[/tex] and [tex]3\sqrt{5}[/tex]
Step-by-step explanation:
Let
x -----> the smaller positive real number
y -----> the larger positive real number
we know that
A positive real number is 2 less than another
so
[tex]x=y-2[/tex]
[tex]y=x+2[/tex] ----> equation A
When 4 times the larger is added to the square of the smaller, the result is 49
so
[tex]4y+x^2=49[/tex] ----> equation B
substitute equation A in equation B
[tex]4(x+2)+x^2=49[/tex]
solve for x
[tex]x^2+4x-41=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\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^2+4x-41=0[/tex]
so
[tex]a=1\\b=4\\c=-41[/tex]
substitute in the formula
[tex]x=\frac{-4\pm\sqrt{4^{2}-4(1)(-41)}} {2(1)}[/tex]
[tex]x=\frac{-4\pm\sqrt{180}} {2}[/tex]
[tex]x=\frac{-4\pm6\sqrt{5}} {2}[/tex]
[tex]x=-2\pm3\sqrt{5}[/tex]
so
The positive real number is
[tex]x=-2+3\sqrt{5}[/tex]
Find the value of y
[tex]y=x+2[/tex]
[tex]y=-2+3\sqrt{5}+2[/tex]
[tex]y=3\sqrt{5}[/tex]