Find the value of k for which the quadratic equation (k-2)x squared+2x(2 k-3)+ 5k-6=0
Answers
Question:
Find the value of k for which the quadratic equation (k-2)x² + 2(2k-3)x + 5k - 6 = 0 has equal roots.
Answer:
k = 1 , 3
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 .
• 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 ;
(k-2)x² + 2(2k-3)x + 5k - 6 = 0
Clearly , we have ;
a = k-2
b = 2(2k-3)
c = 5k-6
We know that ,
The quadratic equation will have equal roots if its discriminant is equal to zero .
=> D = 0
=> [2(2k-3)]² - 4•(k-2)(5k-6) = 0
=> 4(2k-3)² - 4•(5k²-6k-10k+12) = 0
=> 4(4k²-12k+9)² - 4(5k²-16k+12) = 0
=> 4(4k² - 12k + 9 - 5k² + 16k - 12) = 0
=> - k² + 4k - 3 = 0
=> k² - 4k + 3 = 0
=> k² - k - 3k + 3 = 0
=> k(k-1) - 3(k-1) = 0
=> (k-1)(k-3) = 0
=> k = 1 , 3