Find the value of k for which the quadratic equation 9xsquare-3kx+k=0 has equal roots
Answers
b^2-4ac=0
so
(-3)^2-4*9*k=0
9-36k=0
9=36k
k=1/4
Question:
Find the value of k for which the quadratic equation 9x² - 3kx + k = 0 has equal roots.
Answer:
k = 0 , 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 and equal roots if its discriminant is equal to zero .
=> D = 0
=> (-3k)² - 4•9•k = 0
=> 9k² - 4•9•k = 0
=> 9k(k-4) = 0
=> k = 0 , 4