question_answer

Find that value of p for which the quadratic equation \[(p+1){{x}^{2}}-6(p+1)x+3(p+9)=0,p\ne -1\] has equal roots. Hence find the roots of the equation.

Answer:

Given, \[(p+1){{x}^{2}}-6(p+1)x+3(p+9)=0,p\ne -1\]

For equation to have equal roots

\[{{[6(p+1)]}^{2}}-4(p+1).3(p+9)=0\]

\[36{{(p+1)}^{2}}-12(p+1)(p+9)=0\]

\[12(p+1)[3p+3-p-9]=0\]

\[12(p+1)(2p-6)=0\]

\[24(p+1)(p-3)=0\]

\[p=-1\] or \[3\]

So,                           \[p=3\]

As                             \[p\ne -1\]

Now the given equation become

\[4{{x}^{2}}-24x+36=0\]

\[{{x}^{2}}-6x+9=0\]

\[{{x}^{2}}-3x-3x+9=0\]

\[x(x-3)-3(x-3)=0\]

\[(x-3)(x-3)=0\]

\[x=3,3\]

\[\therefore \] Roots are \[3,3\].