Question

find the values of K for which the rroots are real and equal in each of the following equation: 1)Kx(x-2)+6

Answer 1

D = b^2-4ac =0

4k^2 - 4×k×6 =0
k^2 - 6k = 0
k = 6

Answer 2

Here is the answer

Solution

kx(x - 2) + 6 = 0

Therefore,
In standard form ,
We can write it as :-
$\bf{{kx}^{2} - 2kx + 6 = 0}$
Here,
a = k
b = -2k
c = 6

$\bf{ \triangle = {b}^{2} - 4ac}$
$\bf{= { (- 2k)}^{2} - 4(k)(6)}$
$\bf{= {4k}^{2} - 24k}$

Their roots are equal... Given

$\therefore \: \triangle = 0$

$\bf{4k {}^{2} - 24k = 0 }\\ \\ \ \bf{ 4k(k - 6) = 0}$
$\bf{4k = 0 \: \: \: \: or \: \: \: \: k - 6 = 0}$
$\bf {\therefore \: k = 0 \: \: \: or \: \: \: k = 6}$

Now,
k = 0 is unacceptable because if we take value of k = 0, then the equation would have no meaning.

Therefore,
K = 6.

Final answer =>
The value of k is 6.

Thanks!!