Question

Find the values of p for which the quadratic equation 

\[\left( 2p + 1 \right) x^2 - \left( 7p + 2 \right)x + \left( 7p - 3 \right) = 0\] has equal roots. Also, find these roots.

Answer 1

The given quadric equation is \[\left( 2p + 1 \right) x^2 - \left( 7p + 2 \right)x + \left( 7p - 3 \right) = 0\] and roots are real and equal.

Then, find the value of p.

Here, 

\[a = 2p + 1, b = - 7p - 2 \text { and } c = 7p - 3\]

As we know that \[D = b^2 - 4ac\]

Putting the values of  \[a = 2p + 1, b = - 7p - 2 \text { and } c = 7p - 3\].

\[D = \left[ - \left( 7p + 2 \right) \right]^2 - 4\left( 2p + 1 \right)\left( 7p - 3 \right)\]

\[ = (49 p^2 + 28p + 4) - 4\left( 14 p^2 - 6p + 7p - 3 \right)\]

\[ = 49 p^2 + 28p + 4 - 56 p^2 - 4p + 12\]

\[ = - 7 p^2 + 24p + 16\]

The given equation will have real and equal roots, if D = 0

Thus, 

\[- 7 p^2 + 24p + 16 = 0\]

\[\Rightarrow 7 p^2 - 24p - 16 = 0\]

\[ \Rightarrow 7 p^2 - 28p + 4p - 16 = 0\]

\[ \Rightarrow 7p(p - 4) + 4(p - 4) = 0\]

\[ \Rightarrow (7p + 4)(p - 4) = 0\]

\[ \Rightarrow 7p + 4 = 0 \text { or } p - 4 = 0\]

\[ \Rightarrow p = - \frac{4}{7} \text { or } p = 4\]

Therefore, the value of p is 4 or \[- \frac{4}{7}\].

Now, for p = 4, the equation becomes

\[9 x^2 - 30x + 25 = 0\]

\[ \Rightarrow 9 x^2 - 15x - 15x + 25 = 0\]

\[ \Rightarrow 3x(3x - 5) - 5(3x - 5) = 0\]

\[ \Rightarrow (3x - 5 )^2 = 0\]

\[ \Rightarrow x = \frac{5}{3}, \frac{5}{3}\]

for p = \[- \frac{4}{7}\] the equation becomes

\[\left( - \frac{8}{7} + 1 \right) x^2 - \left( - 4 + 2 \right)x + \left( - 4 - 3 \right) = 0\]

\[ \Rightarrow \left( \frac{- 8 + 7}{7} \right) x^2 + 2x - 7 = 0\]

\[ \Rightarrow - \frac{1}{7} x^2 + 2x - 7 = 0\]

\[ \Rightarrow - x^2 + 14x - 49 = 0\]

\[ \Rightarrow x^2 - 14x + 49 = 0\]

\[ \Rightarrow x^2 - 7x - 7x + 49 = 0\]

\[ \Rightarrow x(x - 7) - 7(x - 7) = 0\]

\[ \Rightarrow (x - 7 )^2 = 0\]

\[ \Rightarrow x = 7, 7\]

Hence, the roots of the equation are \[\frac{5}{3} \text{ and } 7\].