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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.

Complete step-by-step answer:

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.

i.e., quadratic equation is

$

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.

Complete step-by-step answer:

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.

i.e., quadratic equation is

$

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.