(3k+1)x²+2(k+1)x+k=0 roots are real and equal
Answers
Answer :
k = 1 , 1/3
Note :
★ The possible values of the variable which satisfy the equation are called its roots or solutions .
★ A quadratic equation can have atmost two roots .
★ The general form of a quadratic equation is given as ; ax² + bx + c = 0
★ If α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ;
• Sum of roots , (α + ß) = -b/a
• Product of roots , (αß) = c/a
★ If α and ß are the roots of a quadratic equation , then that quadratic equation is given as : k•[ x² - (α + ß)x + αß ] = 0 , k ≠ 0.
★ The discriminant , D of the quadratic equation ax² + bx + c = 0 is given by ;
D = b² - 4ac
★ If D = 0 , then the roots are real and equal .
★ If D > 0 , then the roots are real and distinct .
★ If D < 0 , then the roots are unreal (imaginary) .
Solution :
Here ,
The given quadratic equation is ;
(3k + 1)x² + 2(k + 1)x + k = 0
Comparing the given quadratic equation with the general quadratic equation
ax² + bx + c = 0 , we have ;
a = 3k + 1
b = 2(k + 1)
c = k
Also ,
It is given that , the given quadratic equation have real and equal roots .
Thus ,
The discriminant of the given quadratic equation must be equal to zero .
=> D = 0
=> b² - 4ac = 0
=> [ 2(k + 1) ]² - 4×3(k + 1)×k = 0
=> 4(k + 1)² - 4×3k(k + 1) = 0
=> 4(k + 1)•(k + 1 - 3k) = 0
=> (k + 1)(1 - 3k) = 0
=> k = 1 , k = 1/3
Hence , k = 1 , 1/3
Answer: