If 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 $\left(l-m\right){x}^{2}-5\left(l+m\right)x-2\left(l-m\right)=0$ 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 $a{x}^{2}+\mathit{bx}+c=0$  to find the nature of the roots.
Comparing the equations $\left(l-m\right){x}^{2}-5\left(l+m\right)x-2\left(l-m\right)=0$  and $a{x}^{2}+\mathit{bx}+c=0$ , 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 = ${b}^{2}-4\mathit{ac}$, we get
D = $\left[-5{\left(l+m\right)}^{2}-4\left(l-m\right)\left[-2\left(l-m\right)\right]$
Simplifying the expression, we get
D = 25${\left(l+m\right)}^{2}$+8${\left(l-m\right)}^{2}$
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
${\left(l+m\right)}^{2}>0$
Multiplying both sides by 25, we get
25${\left(l+m\right)}^{2}$> 0
Since l−m is not equal to zero, the square of l−m is always positive.
Therefore, we get
${\left(l-m\right)}^{2}>0$ Multiplying both sides by 8, we get
8${\left(l-m\right)}^{2}$> 0
Now, we can observe that 25${\left(l+m\right)}^{2}$> 0 and 8${\left(l-m\right)}^{2}$> 0.
Therefore, we can say that
25${\left(l+m\right)}^{2}$+8${\left(l-m\right)}^{2}$> 0
Substituting D=25${\left(l+m\right)}^{2}$+8${\left(l-m\right)}^{2}$ in the inequation, we get
D > 0
Thus, the discriminant of the equation ${\left(l-m\right)}^{2}-5\left(l+m\right)x-2\left(l-m\right)=0$  is greater than 0.
Therefore, the roots of the equation ${\left(l-m\right)x}^{2}-5\left(l+m\right)x-2\left(l-m\right)=0$ are real and unequal.
Thus, the correct option is option (3).

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