How do you solve #(x-2)² + (-x+6)² = 32#?

1 Answer
Mar 14, 2017

Answer:

#x=4-2sqrt(3)color(white)("XX")orcolor(white)("XX")x=4+2sqrt(3)#

Explanation:

Given
#color(white)("XXX")color(green)((x-2)^2)+color(blue)((-x+6)^2)=color(brown)(32)#

Expanding the left side
#color(white)("XXX")color(green)(x^2-4x+4)+color(blue)(x^2-12x+36)=color(brown)(32)#

Combining like terms
#color(white)("XXX")2x^2-16x+40=color(brown)(32)#

Dividing both sides by #2# to simplify
#color(white)("XXX")x^2-8x+20=16#

Converting to standard quadratic form
#color(white)("XXX")color(red)1x^2color(magenta)(-8x)color(orange)(+4)=0#

With no obvious (at least to me) rational factors:
Apply the quadratic formula
#color(white)("XXX")x=(-color(magenta)b+-sqrt(color(magenta)b^2-4color(red)acolor(orange)c))/(2color(red)a)#

#color(white)("XXXXX")=(8+-sqrt(64-16))/2#

#color(white)("XXXXX")=4+-2sqrt(3)#