How do you find the zeros, real and imaginary, of # y=(x-3)^2+41 # using the quadratic formula?

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Answer 1 · 2017-12-13T03:49:32

#x=3+sqrt(41)i,##3-sqrt(41)i#

Given:

#y=(x-3)^2+41#

Expand #(x-3)^2#.

#y=x^2-6x+9+41#

Simplify.

#y=x^2-6x+50# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=1#, #b=-6#, and #c=50#

The zeroes are the x-intercepts, which are the values of #x# when #y=0#.

Substitute #0# for #y#.

#0=x^2-6x+50#

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

Plug in the known values.

#x=(-(-6)+-sqrt((-6)^6-4*1*50))/(2*1)#

Simplify.

#x=(6+-sqrt(-164))/2#

Prime factorize #164#.

#x=(6+-sqrt(2xx2xx41))/2#

Simplify.

#x=(6+-2sqrt(41)i)/2#

Reduce.

#x=(color(red)cancel(color(black)(6))^3+-color(red)cancel(color(black)(2))^1sqrt(41)i)/color(red)cancel(color(black)(2))^1#

Simplify.

#x=3+-sqrt(41)i#

Values for #x#.

#x=3+sqrt(41)i,##3-sqrt(41)i#