Find the value of p for which the quadratic equation (p+1)x2-6(p+1)x+3(p+9)=0 had equal roots , hence find the roots

d = b2-4ac 
(6p+6)2 -4(p+1)(3p+27) =0
36p2+36 +72p -4(3p2 +30p +27) =0
36p2 +36 +72p-12p2-120p -108 =0
24p2-48p-72 =0
devide by 24
p2-2p-3=0
p2+p-3p-3=0
p(p+1)-3(p+1)=0
(p-3)(p+1)=0
p = 3 or -1

 

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It is wrong
  • -21
Here is the answer

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thanks
 
  • 7
Here is the correct answer!

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it will be helpful to you all

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Hope u find it useful 😄😄

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Show that the equation (p+1)x^2=3+p(2x-1)has distinct roots if p is>=3/2
 
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