Question

please solve the above quadratic equation.

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Answer 1

hope this hlp...........

Answer 2

Given Equation is x^2 - 5x - (p-3)(p+2) = 0

= > x^2 - 5x - (p - 3)(p + 2) = 0

= > x^2 - 5x - (p^2 + 2p - 3p - 6) = 0

= > x^2 - 5x - (p^2 - p - 6)= 0 

= > x^2 - 5x - p^2 + p + 6 = 0

The equation is in the form of ax^2 + bx + c = 0.

a = 1,b = -5,c = -p^2 + p + 6.


We know that :

$x = \frac{-b + \sqrt{b^2-4ac} }{2a}$

x = $\frac{-(-5) + \sqrt{(-5)^2 - 4 * 1 * (-p^2 + p + 6) } }{2 * 1}$
 
   = $\frac{5 + \sqrt{25 - 4(-p^2 + p + 6)} }{2}$
   
   = $\frac{5 + \sqrt{25 - 4p^2 - 4p - 24} }{2}$

   = $\frac{5 + \sqrt{4p^2 - 4p + 1} }{2}$

   = $\frac{5 + \sqrt{(2p - 1)^2} }{2}$

   = $\frac{5 + 2p - 1}{2}$

   = $\frac{2p+4}{2}$

   = $\frac{2(p + 2)}{2}$

   = p + 2.



We know that:

$x = \frac{-b - \sqrt{b^2-4ac} }{2a}$

          = $\frac{-5 - \sqrt{(-5)^2 - 4 * 1 * (-p^2 + p + 6)} }{2 * 1}$

         = $\frac{5 - \sqrt{(2p-1)^2} }{2}$

         = $\frac{5 - (2p-1)}{2}$

         = $\frac{5 - 2p + 1}{2}$

         = $\frac{-2p+6}{2}$

         = $\frac{2(-p+3)}{2}$

         = -p+3.


Hope this helps!