Question

find the value of k ,such that the quadratic equation kx^2-5x+3=0 can be factorised in coincident factors​

Answer 1

Coincident factors means factors having equal roots. So we have to find the values of k for which the equation has two equal roots.

In a quadratic equation, the roots are equal if the discriminant (b² - 4ac) is equal to zero.

Comparing to the standard form of a quadratic equation, ax² + bx +c, we get  here

a = k

b = - 5

c = 3

So, put discriminant (b² - 4ac) = 0

⇒ (-5)² - 4(k)3 = 0

⇒ 25 - 12k = 0

⇒ -12k = -25

⇒ 12k = 25

⇒ k = 25/12

Hence, the value of k for which the equation can be factorised in two equal factors is 25/12

Answer 2

Discriminants :

D = 0

$\boxed{\mathsf{b^2 \:- \:4ac \:=\: 0}}$

Here , b = - 5, a = k and c = 3

Putting these values,

$\mathsf{(\:-\:5)^2\: -\: 4* k *3 \:= \:0}$

25 - 12 k = 0

12k = 25

$\mathsf{k \:= \:{\dfrac{25} {12}}}$

Hence, the value of k is " 25/12".