Find p for which (p+1x*x-6(p+1)x+3(p+q)=0,q is …

According to question we have,

(p + 1)x²- 6(p + 1)x + 3(p + 9) = 0. 
Here, a = p + 1, b = -6(p + 1) and c = 3(p + 9)
We know that, D = b² - 4ac
= [-6(p + 1)]² - 4(p + 1)[3(p + 9)]
= 36(p² + 2p + 1) - (4p + 4)(3p + 27)
 = 36p² + 72p + 36 - (12p² + 108p + 12p + 108)
  = 36p² + 72p + 36 - 12p² - 120p - 108
  = 24p² - 48p - 72
As we know that when quadratic equation has real and equal roots, its discrimnant is equal to zero i.e., 
D = b² - 4ac = 0
$\Rightarrow$ 24p² - 48p - 72 = 0
$\Rightarrow$ p² - 2p - 3 = 0
$\Rightarrow$ p² - 3p + p - 3 = 0
$\Rightarrow$ p(p - 3) + 1(p - 3) = 0
$\Rightarrow$(p - 3)(p + 1) = 0
$\Rightarrow$ p - 3 = 0 or p + 1 = 0
$\Rightarrow$      p = 3 or p = -1
After substituting the values of p in the given equation, the two equations will be of the form
 4x² - 24x + 36 = 0 or 0x² - 0x + 24= 0
Since, 0x² - 0x+24= 0 is not a quadratic equation, therefore we neglect it here.
Now Consider the remaining one i.e.,  4x² - 24x + 36 = 0
$\Rightarrow$ x² - 6x + 9 = 0
$\Rightarrow$ x² - 3x - 3x + 9 = 0
$\Rightarrow$ x(x - 3) - 3(x -3) 0
$\Rightarrow$ (x - 3)(x - 3) = 0
$\Rightarrow$ (x - 3)² = 0
$\Rightarrow$ x - 3 = 0
$\Rightarrow$x = 3

Hence the values of x are 3,3.