Question

For what value of k, the root of quadratic equation: (k + 1) x² – 2(k – 1) x + 1 = 0 Are real and equal

Answer 1

Hey there,

kx^(2)+(k+1)x+1= 0
assume k is not equal to 0
(note that if k = 0 we have only one root or value of x, x = -1)
divide by k we get
or x² + [(k+1)/k] x + 1/k = 0
x² + 2[(k+1)/2k] x + 1/k = 0
or
x² + 2[(k+1)/2k] x + (k+1)² /4k² = - 1/k + (k+1)² /4k²
[ x + (k+1)/2k] ² = [ - 4k + (k+1)² ]/4k²

now for equal roots we have RHS [ - 4k + (k+1)² ]/4k² = 0
or (k+1)² = 4k
k² - 2k +1 = 0 or (k - 1)² = 0
or k = 1

Hope this helps!