Question

Find the value of k in quadratic equation KX(x-2)+6=0,so that they have two equal roots.

Answer 1

$b ^{2} - 4ac = 0$
for equal roots it shuld be equal to zero
ie.

Answer 2

Given:

⇒  $kx(x-2)+6=0$

To find:

Value of k = ?

Solution:

From the given equation, we get

⇒  $kx^2-2kx+6=0$

here,

a = k

b = -2k

c = 6

If they have two equal roots, then

⇒  $b^2-4ac=0$

On substituting the values, we get

⇒  $(-2k)^2-4\times k\times 6=0$

⇒                $4k^2-24k=0$

On adding "24k" both sides, we get

⇒      $4k^2-24k+24k=24k$

⇒                          $4k^2=24k$

⇒                             $k=\frac{24}{4}$

⇒                             $k=6$

Thus the value of k is "6".