Relax

Respuesta :

apologiabiology
we assume that
if xy=0
x and y=0

first make it equal to zero

minus 5 on both sides
4x^2-12x+4=0
undistribute 4 for ease
4(x^2-3x+1)=0
since 4 does not equal zero, and the only thing that will get it to zero is 0,
x^2-3x+1=0
use quadratic equation

if you hahve
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^{2}-4ac} }{2a}[/tex]
1x^2-3x+1=0
a=1
b=-3
c=1

x=[tex] \frac{-(-3)+/- \sqrt{(-3)^{2}-4(1)(1)} }{2(1)}[/tex]
x=[tex] \frac{3+/- \sqrt{9-4} }{2}[/tex]
x=[tex] \frac{3+/- \sqrt{5} }{2}[/tex]
x=[tex] \frac{3+ \sqrt{5} }{2}[/tex] or [tex] \frac{3- \sqrt{5} }{2}[/tex]


aprox
x=2.61803 or 0.381966


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