Find the values of k for which the quadratic equation 9x 2-3kx+4=0 has the real root
Answers
quadratic equation is
9x²-3kx+4=0
on comparing with general eqn
a= 9
b = -3k
c = 4
for real roots
condition is
D ≥0
b²-4ac ≥0
9k²-144≥0
(3k-12)(3k+12)≥0
9(k-4)(k+4)≥0
(k-4)(k+4)≥0
thus k € (-∞,4] U [4,∞)
Question:
Find the value of k for which the quadratic equation 9x² - 3kx + 4 = 0 has real roots.
Answer:
k € [-∞,-4]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 + 4 = 0
Clearly , we have ;
a = 9
b = -3k
c = 4
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•4 ≥ 0
=> 9k² - 9•16 ≥ 0
=> 9•(k² - 16) ≥ 0
=> k² - 16 ≥ 0
=> (k+4)(k-4) ≥ 0
=> k € [-∞,-4]U[4,∞]
Hence,
The required values of k are [-∞,-4]U[4,∞] .