Question

# If the sum and product of roots of a quadratic equation are $- \dfrac{7}{2}$ and $\dfrac{5}{2}$ respectively, then the equation is$\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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Hint: In this particular question use the concept 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 ${x^2}$ and the product of the roots is the ratio of the constant term to the coefficient of ${x^2}$, so use these concepts to reach the solution of the question.

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.
Hence option (a) is the correct answer.

Note:We can also solve this problem directly, if the sum and the product of the roots of the quadratic equation are given then we directly write the quadratic equation which is given as, ${x^2} + \left( { - {\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0$ so simply substitute the given values in this equation and simplify we will get the required quadratic equation.