show that if k is any real number, the equation x^2 + (k+1)x + k = 0 always has real roots. For what value of k are the roots equal?
$\displaystyle \text{descriminant} = {b^2 - 4ac} $
$\displaystyle = {(k+1)^2 - 4(a)(k)} $
$\displaystyle ={k^2+2k+1 - 4k}$
$\displaystyle = {k^2-2k+1} $
For these to be real
$\displaystyle k^2 - 2k + 1 \geq 0 $
$\displaystyle (k-1)^2 \geq 0 $
Is this true for all real numbers ?
For these to be EQUAL
$\displaystyle k^2 - 2k + 1 = 0 $
$\displaystyle (k-1)^2 = 0 $
$\displaystyle k-1 = 0 $
$\displaystyle k = 1 $