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Question
Find the values of p for which the quadratic equation `(p+1)x^2-6(p+1)x+3(p+9)=0` `p≠-11`has equal roots. Hence find the roots of the equation.
Solution
The given equation is` (p+1)x^2-6(p+1)x+3(p+9)=0`
This is of the form `ax^2+bx+c=0 where a=p+1,b=-6(p+1) and c=3(p+9)`
∴` D=b^2-4ac`
=`[-6(p+1)]^2-4xx(p+1)xx3(p+9)`
=`12(p+1)[3(p+1)-(p+9)]`
=`12(p+1) (2p-6)`
The given equation will have real and equal roots if D = 0.
∴ `12(p+1) (2p-6)=0`
⇒`p+1=0 or 2p-6=0`
⇒` p=-1 or p=3`
But, `p≠ -1` (Given)
Thus, the value of p is 3
Putting p =3, the given equation becomes `4x^2-24x+36=0`
4x^2-24x+36=0
⇒ `4(x^2-6x+9)=0`
⇒` (x-3)^2=0`
⇒`x-3=0`
⇒`x=3`
Hence, 3 is the repeated root of this equation.
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