Question

Find the value of k for which the quadratic equation 9x^2-3kx+k=0 has real roots

Answer 1

For the equations having real roots the value of 
D = b²- 4ac = 0 
hence
[3k]² - 4 *9 *k = 0 
9k² - 36k = 0
9k [k-4] = 0
value of k = 0 or k = 4 
Hope this helps 
Thank you

Answer 2

Question:

Find the value of k for which the quadratic equation 9x² - 3kx + k = 0 has real roots.

Answer:

k = [-∞,0]U[4,∞]

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• A quadratic equation has atmost two roots .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

9x² - 3kx + k = 0

Clearly , we have ;

a = 9

b = -3k

c = k

We know that ,

The quadratic equation will have real roots if its discriminant is greater than or equal to zero .

=> D ≥ 0

=> (-3k)² - 4•9•k ≥ 0

=> 9k² - 4•9•k ≥ 0

=> 9k(k-4) ≥ 0

=> k(k-4) ≥ 0

=> k € [-∞,0]U[4,∞]

Hence,

The required values of k are [-∞,0]U[4,∞] .