Question

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

Answer 1

for equal roots discriminant =0
D=√[(-3k)^2-4(9k)]=0
square b. s
9k^2=36k
k(9k)=k(36)
k=0.............................................(1)
9k=36
k=4................................................(2)
k=4,0

Answer 2

Answer:

Hey I Am Ritu Priya

Sonar

Here Is Your Answer

Given quadratic equation is

= 9x² + 3kx + 4 = 0

On comparing with standard form of quadratic equation i.e ax² + bx + c =0,a≠0

Here, a = 9 , b= 3k, c= 4

D(discriminant)= b²-4ac

= (3k)² - 4× 9 ×4

= 9k² - 144

Since, roots of given equation are distinct.  

D > 0.

= 9k² - 144 > 0

= 9(k² - 16) >0

= (k² - 16) >0            (9≠0)

= k² -4²>0

= (k-4) (k+4) >0

= [ a² - b² = (a-b)(a+b)]

= k > 4 and k< -4

Hence, the value of k is k > 4 and k< -4.

THANKS