Math
posted by Emily on .
For what values of k does the function f(x)=(k+1)x^2+2kx+k1 have no zeros? One zero? Two zeros?

You must calculate Discriminant Ä
For quadratic equation:
ax^2+bx+c , Ä=b^24*a*c
If the discriminant is positive, then there are two distinct roots.
If the discriminant is zero, then there is exactly one distinct real root, sometimes called a double root.
If the discriminant is negative, then there are no real roots. Rather, there are two distinct (nonreal) complex roots, which are complex conjugates of each other. 
Ä is Greek letter Delta(letter like triangle).
y=ax^2+bx+c
In your case a=k+1 , b=2k , c=k1
Ä=b^24*a*c
Ä=(2k)^24*(k+1)*(k1)
=4k^24*(k^2+kk1)=4k^24*(k^21)
=4k^24k^2(4)=0+4=4
You equatin have 2 real roots 
In google type "quadratic equation"
Then click on en.wikipedia link which will appered.
On this en.wikipedia page you have all about quadratic equation. 
Thank you!