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# If p, q are the zeroes of the polynomial $p\left( x \right) = 2{x^2} + 5x + k$ satisfying the relation ${p^2} + {q^2} + pq = \dfrac{{21}}{4}$, then what is the possible value of k.

Hint- Here, we will proceed by using the formula for the sum of the roots i.e., $p + q = - \dfrac{b}{a}$ and the product of the roots i.e., $pq = \dfrac{c}{a}$ of any general quadratic equation $a{x^2} + bx + c = 0$ whose roots are p and q.

Given a quadratic polynomial is $p\left( x \right) = 2{x^2} + 5x + k$ having p and q as the roots or zeros of this quadratic polynomial.
Also given, ${p^2} + {q^2} + pq = \dfrac{{21}}{4}{\text{ }} \to {\text{(1)}}$
By equating the given quadratic polynomial with zero, we will get the quadratic equation corresponding to this quadratic polynomial.
$p\left( x \right) = 0 \\ \Rightarrow 2{x^2} + 5x + k = 0{\text{ }} \to {\text{(2)}} \\$
As we know that for any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(3)}}$ having its roots as p and q,
Sum of its roots, $p + q = - \dfrac{b}{a}{\text{ }} \to {\text{(4)}}$
Product of its roots, $pq = \dfrac{c}{a}{\text{ }} \to {\text{(5)}}$
By comparing equations (2) and (3), we get
a = 2, b = 5 and c = k
By putting a = 2, b = 5 and c = k in equations (4) and (5), we get
Sum of its roots, $p + q = - \dfrac{5}{2}{\text{ }} \to {\text{(6)}}$
Product of its roots, $pq = \dfrac{k}{2}{\text{ }} \to {\text{(7)}}$
By adding and subtracting pq in the LHS of the Equation (1) can be rewritten as
$\Rightarrow {p^2} + {q^2} + pq + pq - pq = \dfrac{{21}}{4} \\ \Rightarrow \left[ {{p^2} + {q^2} + 2pq} \right] - pq = \dfrac{{21}}{4} \\$
Using the formula ${\left( {p + q} \right)^2} = {p^2} + {q^2} + 2pq$ in the above equation, we get
$\Rightarrow {\left( {p + q} \right)^2} - pq = \dfrac{{21}}{4}$
Using equations (6) and (7) in the above equation, we get
$\Rightarrow {\left( { - \dfrac{5}{2}} \right)^2} - \left( {\dfrac{k}{2}} \right) = \dfrac{{21}}{4} \\ \Rightarrow \left( {\dfrac{k}{2}} \right) = {\left( { - \dfrac{5}{2}} \right)^2} - \dfrac{{21}}{4} \\ \Rightarrow \dfrac{k}{2} = \dfrac{{25}}{4} - \dfrac{{21}}{4} \\ \Rightarrow k = 2\left( {\dfrac{{25 - 21}}{4}} \right) \\ \Rightarrow k = 2\left( {\dfrac{4}{4}} \right) \\ \Rightarrow k = 2 \\$
Therefore, the required possible value of k is 2.

Note- In this particular problem, we have represented the given relation between the roots of the given quadratic polynomial i.e., ${p^2} + {q^2} + pq = \dfrac{{21}}{4}$ in terms of the sum of roots i.e., (p+q) and product of roots i.e., pq. By equating the polynomial of any degree with zero, we can obtain the equation of that degree corresponding to the given polynomial.