Question

find the value of k for which the equation x2+k(2x+k-1)=0 has real and equal roots

Answer 1

$x2 + 2xk + k2 - k$
as eq. has real and equal root. so
D=b2-4ac.
$where \: a = 1 \: b = 2k \: and \: c = k2 - k \: and \: d = 0$

Answer 2

Given,

An equation: x^2+k(2x+k-1)=0 has real and equal roots.

To find,

The value of k.

Solution,

We can simply solve this mathematical problem using the following process:

Mathematically,

In a quadratic equation: ax^2 + bx + c = 0, the nature of its roots is determined by the value of discriminant D = (b^2-4ac), as follows:

a) if D>0, then real and distinct roots

b) if D=0, then real and equal roots

c) if D<0, then complex and distinct roots

{Statement-1}

The given equation = x^2+k(2x+k-1)

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

According to the given equation,

The value of a = 1

value of b = 2k

value of c = k(k-1)

Now, according to the question;

The given equation has real and equal roots

=> the value of D for the given equation = 0

{according to statement-1}

=> b^2-4ac = 0

=> b^2=4ac

=> (2k)^2 = 4(1)×k(k-1)

=> 4k^2 = 4k^2 - 4k

=> -4k = 0

=> k = 0

Hence, the value of k is equal to 0.