1. ## Using Discriminant

1. Show (k+1)x^2 - 2x - k = 0 has a solution for all values of k.

So, I find discrim, = 4 + 4k^2 + 4k,

Then I find the discrim of that = -48 = negative, meaning no solutions and graph of first discrim is above x-axis.

How does prove that the original has solutions for all k?

2. Show that ax^2 -(a+b)x + b = 0, has solution for all a, b.

Discriminant is a^2 - 2ab + b^2, how do I interpret that?

2. Originally Posted by classicstrings
1. Show (k+1)x^2 - 2x - k = 0 has a solution for all values of k.

So, I find discrim, = 4 + 4k^2 + 4k,
Then I find the discrim of that = -48 = negative, meaning no solutions and graph of first discrim is above x-axis.
How does prove that the original has solutions for all k?

2. Show that ax^2 -(a+b)x + b = 0, has solution for all a, b.

Discriminant is a^2 - 2ab + b^2, how do I interpret that?
Hi, classicstring,

to 1.) Could it be, that there are some restriction for k? Otherwise you have to set
4 + 4k^2 + 4k ≥ 0 and solve for k.

to 2.)

a^2 - 2ab + b^2 = (a-b)² ≥ 0

EB

3. Originally Posted by classicstrings
1. Show (k+1)x^2 - 2x - k = 0 has a solution for all values of k.

So, I find discrim, = 4 + 4k^2 + 4k,
Note that 4k^2 + 4k + 4 >= 0 for all k.

-Dan

4. Hello, classicstrings!

1. Show (k+1)x² - 2x - k .= .0 has a solution for all values of k.

So, I find: .discrim. .= .4k² + 4k + 4
A quadratic has solutions if its discriminant is > 0.

Can our discrimimant ever be negative?

4k² + 4k + 4 has a minimum point at its vertex: .k = -b/2a

Hence, the vertex is at: .k .= .(-4)/(8) .= .

. . and the polynomial equals: .4(-½)² +4(-½) + 4 .= .3

The discriminant's minimum value is 3; hence, it is never negative.

Therefore, the quadratic always has a solution.