For what values of k does the function f(x)=(k+1)x^2+2kx+k-1 have no zeros? One zero? Two zeros?
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.
Thank you!
oop-
Well, let's analyze the options here.
For the function f(x) to have no zeros, it means that the quadratic expression (k+1)x^2 + 2kx + k-1 should never be equal to zero. Since the discriminant determines the number of zeros, we can use that to figure it out.
The discriminant is given by b^2 - 4ac, where a = (k+1), b = 2k, and c = (k-1).
No Zeros: For the discriminant to be negative, we need b^2 - 4ac < 0. Substituting the values, we get:
(2k)^2 - 4(k+1)(k-1) < 0
4k^2 - 4(k^2 - 1) < 0
4k^2 - 4k^2 + 4 < 0
4 < 0
Oh no! 4 is not less than zero, which means there are no values of k that make the function have no zeros. Sorry for the zero zilch nada!
One Zero: For the discriminant to be equal to zero, we need b^2 - 4ac = 0. Substituting the values, we get:
(2k)^2 - 4(k+1)(k-1) = 0
4k^2 - 4(k^2 - 1) = 0
4k^2 - 4k^2 + 4 = 0
4 = 0
Oh dear! 4 is not equal to zero either, which means there are no values of k that make the function have exactly one zero. Looks like zero is MIA!
Two Zeros: For the discriminant to be positive, we need b^2 - 4ac > 0. Substituting the values, we get:
(2k)^2 - 4(k+1)(k-1) > 0
4k^2 - 4(k^2 - 1) > 0
4k^2 - 4k^2 + 4 > 0
4 > 0
Hurray! 4 is indeed greater than zero. That means there are infinite values of k that make the function have two zeros. The possibilities are endless!
So, to sum it up: no zeros are not possible, one zero is out of the equation, and two zeros are the life of the party!
To find the number of zeros of the quadratic function f(x), we can use the discriminant. The discriminant of a quadratic function ax^2+bx+c is given by the formula Δ = b^2 - 4ac.
1. No zeros: If the discriminant is negative (Δ < 0), the quadratic function has no real roots. In this case, there are no values of k for which the function f(x) has no zeros.
2. One zero: If the discriminant is zero (Δ = 0), it means the quadratic function has one real root. In this case, the discriminant can be set to zero and solved for k.
To apply this to the function f(x) = (k+1)x^2 + 2kx + k - 1:
Δ = (2k)^2 - 4(k+1)(k-1)
Simplifying:
Δ = 4k^2 - 4(k^2 - 1)
Δ = 4k^2 - 4k^2 + 4
Δ = 4
Δ = 4 implies that the quadratic function has one real root. Therefore, there is exactly one value of k for which f(x) has one zero.
3. Two zeros: If the discriminant is positive (Δ > 0), it means the quadratic function has two distinct real roots. In this case, again, we solve the discriminant equation for k.
For the function f(x) = (k+1)x^2 + 2kx + k - 1:
Δ > 0 implies 4 > 0, which is always true.
Thus, the function f(x) always has two distinct real roots for any value of k.
To summarize:
- No zeros: There are no values of k for which f(x) has no zeros.
- One zero: f(x) has one zero when the discriminant Δ = 4, leading to a single value of k.
- Two zeros: f(x) always has two zeros for any value of k, as Δ is always greater than zero.
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.
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