Question

find the value of K the quatartic equation Kx(5x-6)+9=0has equal rooys

Answer 1

kx(5x-6)+9=0
5kx^2 -6kx+9=0
here a=5k, b=-6k, c=9
therefore, D=b^2-4ac
given that D =0
so,
D=(-6k)^2-4×5k×9
0=36k^2-180k
36k(k-5)=0
therefore, 36k(k-5)=0
36k=0
k=0
OR
k-5=0
k=5
if we put k=0 then the equation is not possible
therefore, k=5. answer

Answer 2

Given:

Quadratic equation - Kx(5x-6)+9=0

Roots are equal.

To Find:

Find the value of K?

Step-by-step explanation:

• We have the quadratic equation kx(5x−6)+9=0
• We know that when roots are equal the value of D

        $D\geq 0$

• Now Multiply the kx with the inner value and solve the equation, we get:

         $5kx^2-6kx+9=0$

• Now compare the above equation with the general equation

        $5kx^2-6kx+9=0\\\alpha x^2+ \beta x + \gamma =0\\\textrm{So that we have} \ \alpha= 5k, \beta = -6k, \gamma= 9$

• Now put the value of variables here, we get :

       $\beta^2 -4 \times\alpha \times \gamma \\\Rightarrow (-6k)^2- 4 \times 5k \times 9=0\\\Rightarrow 36k^2- 36\times 5k=0\\\Rightarrow 36 k (k-5) =0\\\Rightarrow k=0, k= 5$

Therefore the value of "K" is either "0 or 5".