Question

Find value of k, so that the quadratic equation (k+1)x2 - 2(k - 1)x +1 =
0 has equal roots.​

Answer 1

Step-by-step explanation:

may my answer helpful to you

Answer 2

QUESTION:

The correct question should be:

Find the value of k, so that the quadratic equation $(k+1)x^2 - 2(k - 1)x + 1 = 0$ has equal roots.

ANSWER:

The values of k so that the given equation has equal roots are 0 and 3.

Given,

Quadratic equation: $(k+1)x^2 - 2(k - 1)x + 1 = 0$.

To find,

The value of k, so that the given quadratic equation has equal roots.

Solution,

For any quadratic equation, the discriminant, usually denoted by D, is used to determine whether the equation has real and distinct roots, real and equal roots, or, no real roots.

Consider the quadratic equation

$ax^2+bx+c=0$

Then, the discriminant is given by,

$D=b^2-4ac$

Now,

if D > 0, the equation will have real and distinct roots,

if D = 0, the equation will have real and equal roots, and,

if D < 0, the equation will have no real roots.

For the given equation

$(k+1)x^2 - 2(k - 1)x + 1 = 0$

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

So,

$D = [-2(k - 1)]^2 - 4(k+1)(1)$

$\implies D = 4(k - 1)^2 - 4k-4$

$\implies D = 4(k^2 -2k + 1) - 4k-4$

$\implies D = 4k^2 -8k + 4 - 4k-4$

$\implies D = 4k^2 - 12k$

$\implies D = 4k(k-3)$

It is given that the equation has equal roots. So, D = 0.

$\implies D = 4k(k-3)=0$

$\implies 4k(k-3)=0$

⇒ $4k=0,$ or $k-3=0$

⇒ k = 0, k - 3 = 0

⇒ k = 0, k = 3.

Thus, k has values 0 and 3.

Therefore the values of k so that the given equation has equal roots are 0 and 3.