Find the value o k, for which the quadratic equation (k+4)x square +(k+1)x+1=0 has equal roots.
Answers
Question:
Find the value of k for which the quadratic equation (k+4)x² + (k+1)x + 1 = 0 has equal roots.
Answer:
k = -3 , 5
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+4)x² + (k+1)x + 1 = 0
Clearly , we have ;
a = k+4
b = k+1
c = 1
We know that ,
The quadratic equation will have equal roots if its discriminant is equal to zero .
=> D = 0
=> (k+1)² - 4•(k+4)•1 = 0
=> (k+1)² - 4(k+4) = 0
=> k² + 2k + 1 - 4k - 16 = 0
=> k² - 2k - 15 = 0
=> k² - 5k + 3k - 15 = 0
=> k(k-5) + 3(k-5) = 0
=> (k-5)(k+3) = 0
=> k = -3 , 5