Show that the roots of the equation x^2 - (k+1)x + (k-2) = 0 are distinct and real for all values of k.

Question

Please clearly show and state how to answer the question and explain what to do, thank you! :)

Answer 1

use the quadratic equation to find the discriminant

a=1, b=-(k+1), c=(k-2)

the discriminant is k^2-2k+9

k^2-2k+9 must be greater than 0

k^2-2k+9>0

if k<0, then k^2-2k+9>0

if k=0, then 9>0

if k=1,then 8>0

if k=2, then 9>0

if k>2, then k^2-2k is always positive because k^2-2k=k(k-2)>0 and k-2>0

therefore, if k>2, then k^2-2k+9>0 always

therefore there are 2 distinct real roots

Answer 2

Assuming k is a real number:  Let M = k + 1, then k = 1 - M, and k - 2 can be written as 1 - M - 2 or -M - 1.

 

x² - Mx + (-M - 1) = 0 

 

x² - Mx - M - 1 = 0

 

x2 - 1 - Mx - M = 0   

 

(x - 1)(x + 1) - M(x - 1) = 0   factor by grouping

 

(x - 1)(x + 1 - M) = 0  Back substitute

 

(x - 1)(x + 1 - (k + 1)) = 0

 

(x - 1)(x + 1 - k - 1) = 0

 

(x - 1)(x - k) = 0

 

x = 1 or x = k

 

I'm sure the algebra is solid, but proving it has eluded me....

 

A²