Question

Find the values of k for which the quadratic equation kx(x – 3) + 9 = 0 has real equal

roots.

(a) k = 0 or k = 4 (b) k = 1 or k = 4 (c) k = –3 or k = 3 (d) k = –4 or k = 4​

Answer 1

Answer:

a. k=0 or k=4

Step-by-step explanation:

Simplifying the equation, we get

$kx^{2} - 3kx + 9 = 0$

And by the nature of quadratic equation we know, the roots of a quadratic equation are Real and Equal if Discriminant is Zero.

ie

${b}^{2} - 4ac = 0$

Now comparing, we get

$( - 3k)^{2} - 36k = 0$

$9k^{2} - 36k = 0$

$9k^{2} = 36k$

Option a. k=0 and k=4 satisfies this eqaution

Answer 2

Answer:

Answer:

a. k=0 or k=4

Step-by-step explanation:

Simplifying the equation, we get

And by the nature of quadratic equation we know, the roots of a quadratic equation are Real and Equal if Discriminant is Zero.

ie

Now comparing, we get

Option a. k=0 and k=4

Step-by-step explanation: