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?
Hello, classicstrings!
A quadratic has solutions if its discriminant is > 0.1. Show (k+1)x² - 2x - k .= .0 has a solution for all values of k.
So, I find: .discrim. .= .4k² + 4k + 4
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) .= .-½
. . 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.