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+k-1 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^2-4*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 (non-real) 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=k-1

Ä=b^2-4*a*c

Ä=(2k)^2-4*(k+1)*(k-1)

=4k^2-4*(k^2+k-k-1)=4k^2-4*(k^2-1)

=4k^2-4k^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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