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$\left( a \right)2{x^2} + 7x + 5 = 0$

$\left( b \right)2{x^2} - 7x + 5 = 0$

$\left( c \right)2{x^2} - 7x - 5 = 0$

$\left( d \right)2{x^2} + 7x - 5 = 0$

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Given data:

The sum and product of roots of a quadratic equation are $ - \dfrac{7}{2}$ and $\dfrac{5}{2}$ respectively.

Let the quadratic equation be $a{x^2} + bx + c = 0$, where and a, b, and c are the constant real parameters.

Let the roots of this quadratic equation be P and Q.

Now as we know that in a quadratic equation the sum of the roots is the ratio of the negative times the coefficient of x to the coefficient of ${x^2}$.

So the sum of the roots is,

$ \Rightarrow P + Q = \dfrac{{{\text{ - coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}$

In the above quadratic equation the coefficient of x is b and the coefficient of ${x^2}$ is a.

Therefore, P + Q = $\dfrac{{ - b}}{a}$

Now it is given that the sum of the roots is $ - \dfrac{7}{2}$

Therefore, P + Q = $ - \dfrac{7}{2}$

$ \Rightarrow \dfrac{{ - b}}{a} = - \dfrac{7}{2}$

$ \Rightarrow \dfrac{b}{a} = \dfrac{7}{2}$……………. (1)

Now as we know that in a quadratic equation the product of the roots is the ratio of the constant term to the coefficient of ${x^2}$.

So the product of the roots is,

$ \Rightarrow PQ = \dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}$

Therefore, PQ = $\dfrac{c}{a}$

Now it is given that the product of the roots is $\dfrac{5}{2}$

$ \Rightarrow PQ = \dfrac{5}{2}$

$ \Rightarrow \dfrac{c}{a} = \dfrac{5}{2}$……………….. (2)

Now divide the quadratic equation by a throughout we have,

$ \Rightarrow {x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0$

Now substitutes the value of $\dfrac{b}{a}{\text{ and }}\dfrac{c}{a}$ from equation (1) and (2) in the above equation we have,

$ \Rightarrow {x^2} + \dfrac{7}{2}x + \dfrac{5}{2} = 0$

Now multiply by 2 throughout we have,

$ \Rightarrow 2{x^2} + 7x + 5 = 0$

So this is the required quadratic equation.