I need answers not the link
Pls answer

Dear Student,

The solution to your ( 3^rd ) query has been provided below:

We have polynomial  : x²  - l x  = m  , 

x²  - l x  - m  = 0 ,  Let roots are $\alpha$  and $\beta$
And
we know from relationship between zeros and coefficient . 

Sum of zeros  = $-\frac{\mathrm{Coefficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^2}$
So,
$\alpha$  + $\beta$  = $- \frac{-l}{1}$ = l                                         --- ( 1 )

And , Products of zeros  = $\frac{\mathrm{Const}an\mathrm{t} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} {\mathrm{x}}^2}$
So,
$\alpha$  $\beta$  =  - m                                                      --- ( 2 )

From given condition we get :

$\alpha$  - $\beta$  =  1                                                    --- ( 3 )

Now we add equation 1 and 3 we get :

2$\alpha$ = 1 + l 

$\alpha$ = $\frac{1 + l}{2}$ , Substitute that value in equation 1 we get :
$\frac{1 + l}{2}$ + $\beta$ = l  ,

$\beta$ = l - $\frac{1 + l}{2}$ ,

$\beta$ = $\frac{2 l - 1 - l}{2} = \frac{l - 1}{2}$

Now we substitute values of $\alpha$ and $\beta$ in equation 2 and get  :

$$\begin{gathered} \left(\frac{1 + l}{2}\right)\left(\frac{l - 1}{2}\right) = - m \\ \Rightarrow \frac{1}{4}\left(l + 1\right)\left(l - 1\right) = - m \\ \mathrm{We} \mathrm{know} : {\mathrm{a}}^2 - {\mathrm{b}}^2 = \left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right) , \mathrm{So} \\ \Rightarrow \frac{1}{4}\left(l^2 - 1\right)= - m \\ \Rightarrow l^2 - 1= - 4m \\ \mathbf{\Rightarrow }{\mathit{l}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{ }\mathbf{4}\mathit{m}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Hence}\mathbf{ }\mathbf{proved}\mathbf{ }\mathbf{)} \end{gathered}$$


Hope this information will clear your doubts about topic.

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