How do you find the roots, real and imaginary, of #y= (x-3)^2+2 # using the quadratic formula?

1 Answer
Nov 29, 2015

Answer:

Expand the expression on the right side into standard form, then apply the quadratic formula.
In this case there will be two imaginary roots #2+-isqrt(5)#

Explanation:

#y=(x-3)^2+2#

#rArr y= 1x^2+(-4)x+9#

which is an example of quadratic standard form: #y=ax^2+bx+c#
for which the roots are given by the quadratic formula
#color(white)("XXX")x= (-b+1sqrt(b^2-4ac))/(-2a)#

Which for our example becomes:
#color(white)("XXX")x= (-(-4)+-sqrt((-4)^2-4(1)(9)))/(2(1))#

#color(white)("XXXX")= (4+-sqrt(16-36))/2#

#color(white)("XXXX")=(4+-2sqrt(-5))/2#

#color(white)("XXXX")=2+-isqrt(5)#