x^2 = -10
x = ±√-10 = ±i√10
two imaginary or complex roots.
(in this case it was easier to find the actual roots when determining the "nature of the roots".
the value of b^2 - 4ac would have been
0 - 4(1)(10) or -40 , so you would have two imaginary roots)
x^2+10=0
Choose the nature of the solutions of the equation
It is 2 real solutions
1 answer