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

1 Answer
Mar 2, 2018

Answer:

#t=4i#
#t=-4i#

Explanation:

A quadratic expression would be:

#ax^2+bx+c=0#

where #a!=0#, but #b# and #c# can equal zero.

Here, #b=0#. The expression can be rewritten as:

#1t^2+0t+16=0#

Now, we can input the values into the quadratic formula:

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

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

#t=(+-sqrt(-64))/2#

#t=(+-8i)/2#

where #i=sqrt-1#

now, #t# has #2# possible solutions:

1) #t=(+8i)/2#

#t=4i#

2) #t=(-8i)/2#

#=-4i#

The roots are imaginary, and so are the answers.

Another thing to be noted is that according to the Conjugate Pairs Theorem, if #(a+bi)# is the root of a polynomial, #(a-bi)# is also a root, and vice versa. The above fits this rule.