Question #8ce4f

2 Answers
Mar 30, 2017

Answer:

#x=(5+-sqrt(5))/2#

Explanation:

Note that
#color(white)("XXX")color(orange)((x-1)^2)=color(orange)(x^2-2x+1)#

So
#color(orange)((x-1)^2)color(magenta)(-3x+4)=0#
is equivalent to
#color(orange)(x^2-2x+1)color(magenta)(-3x+4)=0#
or
#x^2-5x+5=0#
or
#color(blue)1x^2color(red)(-5)x+color(green)5=0#

Using the quadratic formula:
#color(white)("XXX")#a quadratic of the form #color(blue)ax^2+color(red)bx+color(green)c=0#
#color(white)("XXX")#has solutions #x=(-color(red)b+-sqrt(color(red)(b)- 4color(blue)acolor(green)c))/(2color(blue)a)#

So #color(blue)1x^2color(red)(-5)x+color(green)5=0#

has solutions #x= (+color(red)5+-sqrt((color(red)(-5))^2-4 * color(blue)1 * color(green)5))/(2 * color(blue)1)#

#color(white)("XXXXXXXX")=(5+-sqrt(5))/2#

Mar 30, 2017

Answer:

#x=5/2 +- sqrt5 /2#

Explanation:

Let x-1=y so that the given eq becomes #y^2 -3y-3+4=0#

Or, #y^2 -3y+1=0#

Using the quadratic formula, it would be y=#(3+-sqrt(9-4))/2#

Or, #y=3/2 +- sqrt5 /2#

So that #x=1+3/2 +-sqrt5/2= 5/2 +- sqrt5 /2#