How do you solve #x^2+16=0# using the quadratic formula?

4 Answers
Feb 22, 2018

Answer:

If #x^2+16=0#

#x=+-sqrt(-16)#

Explanation:

If #ax^2+bx+c=0, #
then
#x=1/(2a)(-b+-sqrt(b^2-4ac))#

For #x^2+16=0#

#a=1, b=0, c=16#
Substituting

#x=1/(2xx1)(-0+-sqrt(0^2-4xx1xx16))#

#=1/2xx+-sqrt(-64)#

#=+-sqrt(-16)#

Feb 22, 2018

Answer:

#+-4i#

Explanation:

The quadratic formula is given by:

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

The equation in standard form is

#x^2+0x+16#

and #a=1,b=0,c=16#

Plug in:

#(-0+-sqrt(0^2-4*1*16))/(2*1)#

#(0+-sqrt(-64))/(2)#

The number under the square root is negative, so there are no real solutions.

Finding the imaginary solutions (#i=sqrt(-1)#):

#(+-sqrt(64*-1))/2#

#(+-sqrt(64)*sqrt(-1))/2#

#(+-8i)/2#

#+-4i#

Feb 22, 2018

Answer:

#x=+-4i#

Explanation:

#"An alternative approach"#

#x^2+16=0#

#rArrx^2=-16#

#"this has no real solutions but will have 2 complex"#
#"solutions"#

#color(blue)"take square root of both sides"#

#rArrx=+-sqrt(-16)larrcolor(blue)"note plus or minus"#

#rArrx=+-sqrt(16xx-1)=+-4i#

Feb 23, 2018

Answer:

#x=(0±sqrt(−64))/2#

Explanation:

#x=(−b±sqrt(b^2−4ac))/(2a)#

#x=(−(0)±sqrt((0)^2−4(1)(16)))/(2(1))#

#x=(0±sqrt(−64))/2#