Question

-x^2-9x-3=0 find number of real roots

Answer 1

like my coefficient of x^2 to be positive
so multiply both sides by -1
x^2+9x+3=0
so we have
\[\text{ discrimnant } =b^2-4ac=9^2-4(1)(3)=81-12=69>0 => \text{ 2 real roots}\]

Answer 2

If you don't actually need to know what the roots are, just how many, the easiest way is to take the determinant? I think that's the word...anyway or you can use -b/2a to find the vertex. So It'd look like this:
9/-2 or -4.5. That would be the x coordinate of your vertex of the graph where the x intercepts are your roots. Since the a term is negative, the parabola opens down. So if your vertex is under the x intercept, there are no real roots. So now, plug in -4.5 to your equation and find the y value. Nevermind, no reason to finish...you have your answer right above this! Listen to myininaya :)

Answer 3

lol