Find the Values of p for Which the Quadratic Equation Has Equal Roots. Also, Find These Roots.

Question:

\( (2p+1)x^2-(7p+2)x+(7p-3)=0 \)

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

Solution

For equal roots, the discriminant must be zero.

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

Here,

\( a=2p+1,\quad b=-(7p+2),\quad c=7p-3 \)

Therefore,

\( (7p+2)^2-4(2p+1)(7p-3)=0 \)

\( 49p^2+28p+4-(56p^2+4p-12)=0 \)

\( -7p^2+24p+16=0 \)

\( 7p^2-24p-16=0 \)

\( (7p+4)(p-4)=0 \)

\( p=-\frac{4}{7} \quad \text{or} \quad p=4 \)

When \(p=4\):

\( 9x^2-30x+25=0 \)

\( (3x-5)^2=0 \)

\( x=\frac{5}{3},\frac{5}{3} \)

When \(p=-\frac{4}{7}\):

\( -\frac{1}{7}x^2+2x-7=0 \)

\( x^2-14x+49=0 \)

\( (x-7)^2=0 \)

\( x=7,7 \)

\( \boxed{p=4 \text{ or } p=-\frac{4}{7}} \)

For \(p=4\), the equal roots are \( \frac{5}{3},\frac{5}{3} \)

For \(p=-\frac{4}{7}\), the equal roots are \( 7,7 \)

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