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.

 

Answer 1

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.