Respuesta :
n is 4 more than square of half of n
[tex]n=4+ (\frac{n}{2})^2[/tex]
[tex]n=4+ \frac{n^2}{4}[/tex]
times both sides by 4
4n=16+n²
minus 4n both sides
0=n²-4n+16
gots to use quadratic formula
for an²+bn+c=0
n=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
1n²-4n+16=0
n=[tex] \frac{-(-4)+/- \sqrt{(-4)^2-4(1)(16)} }{2(1)} [/tex]
n=[tex] \frac{4+/- \sqrt{16-64} }{2} [/tex]
n=[tex] \frac{4+/- \sqrt{-48} }{2} [/tex]
uh oh, we got imaginary numbers
if we simplify we get n=2+2i√3 or n=2-2i√3
that can't be right
maybe it's
[tex]n=4+\frac{n^2}{2}[/tex]
times both side sby 2
2n=8+n²
minus 2n
0=n²-2n+8
factor
0=(n-4)(n+2)
st to zero
n-4=0
n=4
n+2=0
n=-2
the number is -2 or 4
if it be [tex]n=4+ (\frac{n}{2})^2[/tex] then n=2+2i√3 or 2-2i√3
if it be [tex]n=4+ \frac{n^2}{2}[/tex] then n=-2 or 4
[tex]n=4+ (\frac{n}{2})^2[/tex]
[tex]n=4+ \frac{n^2}{4}[/tex]
times both sides by 4
4n=16+n²
minus 4n both sides
0=n²-4n+16
gots to use quadratic formula
for an²+bn+c=0
n=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
1n²-4n+16=0
n=[tex] \frac{-(-4)+/- \sqrt{(-4)^2-4(1)(16)} }{2(1)} [/tex]
n=[tex] \frac{4+/- \sqrt{16-64} }{2} [/tex]
n=[tex] \frac{4+/- \sqrt{-48} }{2} [/tex]
uh oh, we got imaginary numbers
if we simplify we get n=2+2i√3 or n=2-2i√3
that can't be right
maybe it's
[tex]n=4+\frac{n^2}{2}[/tex]
times both side sby 2
2n=8+n²
minus 2n
0=n²-2n+8
factor
0=(n-4)(n+2)
st to zero
n-4=0
n=4
n+2=0
n=-2
the number is -2 or 4
if it be [tex]n=4+ (\frac{n}{2})^2[/tex] then n=2+2i√3 or 2-2i√3
if it be [tex]n=4+ \frac{n^2}{2}[/tex] then n=-2 or 4