A quadratic expression would be:
ax^2+bx+c=0ax2+bx+c=0
where a!=0a≠0, but bb and cc can equal zero.
Here, b=0b=0. The expression can be rewritten as:
1t^2+0t+16=01t2+0t+16=0
Now, we can input the values into the quadratic formula:
t=(-b+-sqrt(b^2-4ac))/(2a)t=−b±√b2−4ac2a
t=(-0+-sqrt(0^2-4*1*16))/(2*1)t=−0±√02−4⋅1⋅162⋅1
t=(+-sqrt(-64))/2t=±√−642
t=(+-8i)/2t=±8i2
where i=sqrt-1i=√−1
now, tt has 22 possible solutions:
1) t=(+8i)/2t=+8i2
t=4it=4i
2) t=(-8i)/2t=−8i2
=-4i=−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)(a+bi) is the root of a polynomial, (a-bi)(a−bi) is also a root, and vice versa. The above fits this rule.
