Find the Value of p for Which the Quadratic Equation Has Equal Roots and Also Find the Roots

Question:

\( (p+1)x^2 – 6(p+1)x + 3(p+9) = 0,\; p \ne -1 \)

Find the value of \(p\) for which the equation has equal roots. Also, find the roots.

Solution

For equal roots, the discriminant must be zero.

\( D=b^2-4ac=0 \)

Here,

\( a=p+1,\quad b=-6(p+1),\quad c=3(p+9) \)

Therefore,

\( [-6(p+1)]^2-4(p+1)\cdot3(p+9)=0 \)

\( 36(p+1)^2-12(p+1)(p+9)=0 \)

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

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

\( (p+1)(p-3)=0 \)

\( p=-1 \quad \text{or} \quad p=3 \)

Since \( p\ne -1 \),

\( p=3 \)

Substituting \( p=3 \),

\( 4x^2-24x+36=0 \)

\( x^2-6x+9=0 \)

\( (x-3)^2=0 \)

\( x=3 \)

Equal roots: \( 3,\;3 \)

The equation has equal roots for

\( \boxed{p=3} \)

Corresponding equal roots are:

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