How do you find all the real and complex roots of #x^2 + 8x + 25 = 0#?

2 Answers
Jan 10, 2016

Answer:

#x=-4+-3i#

Explanation:

Complete the square.

#(x^2+8x+16)+25-16=0#

#(x+4)^2+9=0#

#(x+4)^2=-9#

#sqrt((x+4)^2)=sqrt(-9)#

#x+4=+-3i#

#x=-4+-3i#

Jan 10, 2016

Answer:

Roots (both complex) are at #x=-4+3i# and #x=-4-3i#

Explanation:

The quadratic formula tells us that for a quadratic equation in the form:
#color(white)("XXX")ax^2+bx+c=0#
the roots are given by:
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

For the given example: #x^2+8x+25=0#
#color(white)("XXX")a=1, b=8, and c=25#

So the roots are:
#color(white)("XXX")x=(-8+-sqrt(8^2-4(1)(25)))/(2(1))#

#color(white)("XXXX")=(-8+-sqrt(-36))/2#

#color(white)("XXXX")=(-8+-sqrt(36)*sqrt(-1))/2#

#color(white)("XXXX")=-4+-3i#