Question

The equation (l-m/2)x^2-(l+m/2)x+m=0 has got two values of satisfy the equation

Answer 1

Please see the attachment

Answer 2

Given :

The quadratic equation is

($\dfrac{l-m}{2}$  ) x² - (  $\dfrac{l+m}{2}$) x + m = 0

To Find :

The two roots satisfying this equation

Solution :

The standard quadratic equation is  a x² + b x + c = 0

Let  α , β be roots of this equation

So, sum of roots = $\dfrac{-b}{a}$

      product of roots = $\dfrac{c}{a}$

And  x = $\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

So,    from the given equation  , a =   $\dfrac{l-m}{2}$     , b =  $\dfrac{l+m}{2}$  , c = m

i.e  b² - 4 a c =  ( $\dfrac{l+m}{2}$ )² - 4 × ($\dfrac{l-m}{2}$    ) × m

                    =  $\dfrac{l^{2}+m^{2}+2lm }{4}$ - 2 m ( l - m )

                    = $\dfrac{l^{2} + m^{2} + 2 l m - 8 m l + 8 m^{2} }{4}$

                    = $\dfrac{9m^{2}+l^{2}-6lm }{4}$

                    = $\dfrac{(l - 3m)^{2} }{4}$

And   - b = -  ( - ( $\dfrac{l+m}{2}$ ) )

So,     x =  $\dfrac{\dfrac{l+m}{2}\pm \dfrac{l-3m}{2}}{l-m}$

i.e    x = 1 ,  $\dfrac{2m}{l-m}$

The roots  =   1  ,  $\dfrac{2m}{l-m}$

Hence, The roots of equation is    1  ,  $\dfrac{2m}{l-m}$    Answer