Posted by Emily on Friday, October 22, 2010 at 7:22pm.
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, Friday, October 22, 2010 at 9:29pm
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, Friday, October 22, 2010 at 9:39pm
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, Friday, October 22, 2010 at 9:44pm
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, Friday, October 22, 2010 at 10:33pm
Thank you!
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