Question

Solve the following given question:
Find the values \[p\] 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

Hint: This is the given quadratic equation. We find the discriminant of the given quadratic equation to know if the roots are real, equal or imaginary. Since it is given that the equation has real roots, we substitute the discriminant value for equal roots and find the value of the given variable. We then substitute the found variable, in the given question of the roots to get the final solution.

Complete step-by-step solution:
In the given question, it is stated that the roots are equal.
Equal roots imply that the discriminant \[D\] should be equal to zero.
\[ \Rightarrow D = 0\]
The discriminant is given by the formula-
\[D = {b^2} - 4ac\]
Comparing the given equation with the standard form of the quadratic equation, that is \[a{x^2} + bx + c = 0\], we get;
\[a = 2p + 1\], \[b = - \left( {7p + 2} \right)\], and \[c = 7p - 3\]
Substituting the values of \[a,b\] and \[c\], in the discriminant formula, we get:
\[ \Rightarrow {\left( {7p + 2} \right)^2} - \left[ {4\left( {2p + 1} \right)\left( {7p - 3} \right)} \right] = 0\]
Expanding the above equation into simpler terms, we get:
\[ \Rightarrow 49{p^2} + 28p + 4 - \left[ {4\left( {14{p^2} - 6p + 7p - 3} \right)} \right] = 0\]
Rearranging and simplifying the terms, we get:
\[ \Rightarrow 49{p^2} + 28p + 4 - 56{p^2} - 4p + 12 = 0\]
\[ \Rightarrow - 7{p^2} + 24p + 16 = 0\]
Taking the negative sign common and shifting it to the right-hand side, we get:
\[ \Rightarrow 7{p^2} - 24p - 16 = 0\]
Now, to find the value of \[p\], we need to use the quadratic equation solving method so we could find the value of \[p\].
\[ \Rightarrow 7{p^2} - 28p + 4p - 16 = 0\]
Taking the common terms out,
\[ \Rightarrow 7p\left( {p - 4} \right) + 4\left( {p - 4} \right) = 0\]
Yet again, taking the common terms out and rearranging, we get:
\[ \Rightarrow \left( {p - 4} \right)\left( {7p + 4} \right) = 0\]
We get two values of \[p\] from the above equation.
\[ \Rightarrow p = 4\]or \[p = \dfrac{{ - 4}}{7}\]
Therefore, the values of \[p\] for which the roots are equal are \[4, - \dfrac{4}{7}\]
Now, substituting these values of \[p\] in the question, we get the roots of it.
For \[p = 4\],
\[\left( {2p + 1} \right){x^2} - \left( {7p + 2} \right)x + \left( {7p - 3} \right) = 0\]
\[\left( {2\left( 4 \right) + 1} \right){x^2} - \left( {7\left( 4 \right) + 2} \right)x + \left( {7\left( 4 \right) - 3} \right) = 0\]
Simplifying the equation;
\[9{x^2} - 30x + 25 = 0\]
Now, to find the value of \[x\] from the above equation, we write the equation in a perfect square form;
\[9{x^2} - 30x + 25\] can also be written as, \[{\left( {3x - 5} \right)^2}\]
\[ \Rightarrow {\left( {3x - 5} \right)^2} = 0\]
\[x = \dfrac{5}{3},\dfrac{5}{3}\]
For \[p = - \dfrac{4}{7}\]:
\[\left( {2p + 1} \right){x^2} - \left( {7p + 2} \right)x + \left( {7p - 3} \right) = 0\]
\[\left( {2\left( { - \dfrac{4}{7}} \right) + 1} \right){x^2} - \left( {7\left( { - \dfrac{4}{7}} \right) + 2} \right)x + \left( {7\left( { - \dfrac{4}{7}} \right) - 3} \right) = 0\]
Simplifying the equation, we get
\[\left( { - \dfrac{8}{7} + 1} \right){x^2} - \left( { - 4 + 2} \right)x + \left( { - 4 - 3} \right) = 0\]:
Add and subtract the terms,
\[ \Rightarrow - \dfrac{1}{7}{x^2} + 2x - 7 = 0\]
Multiplying the terms by $\left( { - 7} \right)$ we get,
\[ \Rightarrow {x^2} - 14x + 49 = 0\]
Factorize that,
\[ \Rightarrow {\left( {x - 7} \right)^2} = 0\]
Hence,
\[ \Rightarrow x = 7,7\]

$\therefore $ The roots of the equation are equal, that is \[7,7\] or \[\dfrac{5}{3},\dfrac{5}{3}\]

Note: The quadratic equations are solved by the factorization method because the roots are real and equal. In order to find the variable in the first case, there is another quadratic equation you need to solve, to find the value of the given variable. In the second set of quadratic equations, you have to transform the equation in the form of the perfect square to find the roots of the equation because given that the roots are equal.