
If the sum and product of roots of a quadratic equation are $ - \dfrac{7}{2}$ and $\dfrac{5}{2}$ respectively, then the equation is
$\left( a \right)2{x^2} + 7x + 5 = 0$
$\left( b \right)2{x^2} - 7x + 5 = 0$
$\left( c \right)2{x^2} - 7x - 5 = 0$
$\left( d \right)2{x^2} + 7x - 5 = 0$
Answer
577.2k+ views
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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Who was the first woman to receive Bharat Ratna?

Write a letter to the principal requesting him to grant class 10 english CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Discuss the main reasons for poverty in India

