Question

Find the value of k for which the equation 9r square +3kr+4=0has real roots

Answer 1

Answer:

Attached snap

Answer 2

Question:

Find the value of k for which the equation 9r² + 3kr + 4 = 0 has real roots.

Answer:

k € (-∞ , -4) U (4 , ∞)

OR

{ k € R - [-4 , 4] }

Solution:

The given quadratic equation is;

9r² + 3kr + 4 = 0.

To find : The value of k for which ,the given quadratic equation has real roots.

We know that ,

A quadratic equation has real roots if its discriminant is greater than or equal to zero.

Thus,

=> D ≥ 0

=> b² - 4ac ≥ 0

=> (3k)² - 4×9×4 ≥ 0

=> 9k² - 9×16 ≥ 0

=> 9( k ² - 16 ) ≥ 0

=> k² - 16 ≥ 0

=> (k - 4)(k + 4) ≥ 0 ---------(1)

Critical points for the above inequation :

k = -4 , 4

Sign scheme for inequation-(1)

+ – +

–∞ <-----------( – 4)-----------( 4 )------------> ∞

The solution set for inequation-(1) is ;

(-∞ , -4) U (4 , ∞)

OR

R – [-4 , 4]

Hence,

The given quadratic polynomial to have real roots , k € (-∞ , -4) U (4 , ∞)

In other way , we can say ;

k € R – [-4 , 4]