MathematicsIf l,m,n are real, and l≠m, then the roots of the equation l-mx2-5l+mx-2l-m=0 are

If l,m,n are real, and l≠m, then the roots of the equation l-mx2-5l+mx-2l-m=0 are


  1. A
    Real and equal
  2. B
    Complex
  3. C
    Real and unequal
  4. D
    None of these 

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    Solution:

    We need to find the nature of roots of the given quadratic equation.
    We will use the formula for discriminant of a quadratic equation ax2+bx+c=0  to find the nature of the roots.
    Comparing the equations l-mx2-5l+mx-2l-m=0  and ax2+bx+c=0 , we get a = l−m, b = −5(l+m), and c = −2(l−m)
    Substituting a=l−m, b = −5(l+m), and c = −2(l−m) in the formula D = b2-4ac, we get
    D = [-5(l+m)2-4(l-m)[-2l-m]
    Simplifying the expression, we get
    D = 25(l+m)2+8(l-m)2
    We need to check whether this discriminant is more than 0, less than 0, or equal to 0.
    It is given that l ≠ m.
    Therefore, l−m ≠ 0.
    We know that the square of any real number is always positive.
    Therefore, we get
    (l+m)2>0
    Multiplying both sides by 25, we get
     25(l+m)2> 0
    Since l−m is not equal to zero, the square of l−m is always positive.
    Therefore, we get
    (l-m)2>0 Multiplying both sides by 8, we get
    8(l-m)2> 0
    Now, we can observe that 25(l+m)2> 0 and 8(l-m)2> 0.
    Therefore, we can say that
    25(l+m)2+8(l-m)2> 0
    Substituting D=25(l+m)2+8(l-m)2 in the inequation, we get
    D > 0
    Thus, the discriminant of the equation (l-m)2-5l+mx-2l-m=0  is greater than 0.
    Therefore, the roots of the equation (l-m)x2-5l+mx-2l-m=0 are real and unequal.
    Thus, the correct option is option (3).
     
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