How do you find the roots, real and imaginary, of #y= x^2-6x+(x-4)^2 # using the quadratic formula?

1 Answer
Mar 16, 2016

Answer:

real roots: #x=(7+-sqrt(17))/2#
imaginary roorts: none

Explanation:

#1# Start by converting the equation into standard form.

#y=x^2-6x+(x-4)^2#

#y=x^2-6x+(color(crimson)x# #color(blue)(-4))(color(orange)x# #color(teal)(-4))#

#y=x^2-6x+(color(crimson)x(color(orange)x)# #color(crimson)(+x)(color(teal)(-4))# #color(blue)(-4)(color(orange)x)# #color(blue)(-4)(color(teal)(-4)))#

#y=x^2-6x+(x^2-4x-4x+16)#

#y=x^2-6x+(x^2-8x+16)#

#y=x^2+x^2-6x-8x+16#

#y=color(brown)2x^2# #color(turquoise)(-14)x# #color(violet)(+16)#

#2#. Identify the #color(brown)a,color(turquoise)b,# and #color(violet)c# values of the quadratic equation. Then plug the values into the quadratic formula to solve for the roots.

#color(brown)(a=2)color(white)(XXXXX)color(turquoise)(b=-14)color(white)(XXXXX)color(violet)(c=16)#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(color(turquoise)(-14))+-sqrt((color(turquoise)(-14))^2-4(color(brown)2)(color(violet)(16))))/(2(color(brown)2))#

#x=(14+-sqrt(196-128))/4#

#x=(14+-sqrt(68))/4#

#x=(14+-2sqrt(17))/4#

#x=(2(7+-sqrt(17)))/(2(2))#

#x=(color(red)cancelcolor(black)2(7+-sqrt(17)))/(color(red)cancelcolor(black)2(2))#

#color(green)(|bar(ul(color(white)(a/a)x=(7+-sqrt(17))/2color(white)(a/a)|)))#

#:.#, the real roots are #x=(7+-sqrt(17))/2#.