MathematicsFind ‘m’ so that the roots of the equation (4+m)x2+(m+1)x+1= 0 may be equal.

Find ‘m’ so that the roots of the equation (4+m)x2+(m+1)x+1= 0 may be equal.


  1. A
    5,-3
  2. B
    5,3
  3. C
    4,2
  4. D
    1,-5 

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    Solution:

    The given equation is:
     (4+m)x2+(m+1)x+1= 0
    The given equation is identical to the general form of a quadratic equation
     ax2+bx+c = 0
    Here we have,
     a = (4+m), b = (m+1), c = 1
    Now the quadratic formula is -b±b2-4ac 2a
    Putting the respective values of a,b and c we get:
     x=-m+1±m+12-4×4+m×1 2×4+m
     x=-m-1±m2+2m+1-16-4m8+2m
    Applying formula(2)i.e.$a+b2=a2+2ab+b2$
     x=-m-1±m2-2m-158+2m
    Let x1=-m-1+m2-2m-158+2m   and   x2=-m-1-m2-2m-158+2m
    According to the question we want to have x1=x2
     -m-1+m2-2m-158+2m =-m-1-m2-2m-158+2m
    As the denominators are same we can cancel then and write,
     -m-1+m2-2m-15=-m-1-m2-2m-15
    On cancellation of similar terms on both the sides of the equation we gwt,
     m2-2m-15=-m2-2m-15
     2m2-2m-15=0
     m2-5m+3m-15=0
     m(m-5)+3(m-5)=0
     (m+3)(m-5)=0
    Now we have two possibilities that are: (m+3)=0 or (m-5)=0
    So m=-3 or m=5
     
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