Question

16. Find k so that the quadratic equation (k-1)X-2(k+1)x+ 1=0 where the eqn have equal roots​

Answer 1

Step-by-step explanation:

hope this answer will help you

Answer 2

Given in the question

1. roots are equal
2. $(k - 1) {x}^{2} - 2(k - 1)x + 1 = 0$

To find :

value of k

solution :

we know that when roots are equal then Discriminant(D) is zero

formula to be used

${(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy$

$D = {b - 4ac}^{2}$

here ,

• b is the cofficent of x = -2(k - 1)
• a is the cofficent of x^2 = (k - 1)
• c is the constant term = 1

$D = {b}^{2} - 4ac = 0$

$D = {( - 2)}^{2} {(k - 1)}^{2} - 4(k - 1)1 = 0$

$4( {k}^{2} + 1 - 2k) - 4(k - 1) = 0$

${k}^{2} + 1 - 2k - k + 1 = 0$

${k}^{2} + 2 - 3k = 0$

$(k - 2)(k - 1) = 0$

$k = 2,1$