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?

Math 
Bosnian,
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. 
Math 
Bosnian,
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 
Math 
Bosnian,
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. 
Math 
Emily,
Thank you!