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

1 Answer
Mar 2, 2018

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.