For what values of k does the function f(x)=(k+1)x^2+2kx+k-1 have no zeros? One zero? Two zeros?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

Thank you!

Answer 5

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