
If the sum of the roots of the equation ${{x}^{2}}+px+q=0$ is 3 times their difference, then
A. $2{{p}^{2}}=9q$
B. $2{{p}^{2}}=9p$
C. ${{(p+q)}^{2}}=8pq$
D. None
Answer
163.2k+ views
Hint: In this question, we are to find the relation between the roots of the given quadratic equation. For this, we use the sum and product of the roots of a quadratic equation formula. By simply applying the coefficients of the given equation into the formulae, we get the required expression.
Formula UsedConsider a quadratic equation $a{{x}^{2}}+bx+c=0$.
Then, its roots are $\alpha $ and $\beta $. So, the sum of the roots and the product of the roots is formulated as:
$\alpha +\beta =\dfrac{-b}{a};\alpha \beta =\dfrac{c}{a}$
Complete step by step solution:The given quadratic equation is ${{x}^{2}}+px+q=0$ and its roots are $\alpha $ and $\beta $.
So, comparing the given equation with the standard equation, we get
$a=1;b=p;c=q$
Then, the sum of the roots of the given equation is obtained as
$\alpha +\beta =\dfrac{-p}{1}=-p\text{ }...(1)$
And the product of the roots of the given equation is
$\alpha \beta =\dfrac{q}{1}=q\text{ }...(2)$
It is given that the sum of the roots of the equation is 3 times their difference. Thus, we can write
$(\alpha +\beta )=3(\alpha -\beta )$
On simplifying, we get
$\begin{align}
& \Rightarrow \alpha +\beta =3(\alpha -\beta ) \\
& \Rightarrow \alpha +\beta =3\alpha -3\beta \\
& \Rightarrow 2\alpha =4\beta \\
& \Rightarrow \alpha =2\beta \text{ }...(3) \\
\end{align}$
On substituting (3) in (1), we get
$\begin{align}
& \alpha +\beta =-p \\
& \Rightarrow 2\beta +\beta =-p \\
& \Rightarrow 3\beta =-p \\
& \Rightarrow \beta =\dfrac{-p}{3}\text{ }...(4) \\
\end{align}$
On substituting (3) in (2), we get
$\begin{align}
& \alpha \beta =q \\
& \Rightarrow (2\beta )\beta =q \\
& \Rightarrow 2{{\beta }^{2}}=q\text{ }...(5) \\
\end{align}$
Then, substituting (4) in (5), we get
$\begin{align}
& 2{{\beta }^{2}}=q \\
& \Rightarrow 2{{\left( \dfrac{-p}{3} \right)}^{2}}=q \\
& \Rightarrow \dfrac{2{{p}^{2}}}{9}=q \\
& \therefore 2{{p}^{2}}=9q \\
\end{align}$
Thus, this is the required expression.
Option A is correct
Note: Here, we need to remember the sum and product formulae. So, we can substitute these values in the given relation to find an expression. And be sure with the signs in the formulae. Otherwise, the entire calculation may go in the wrong way.
Formula UsedConsider a quadratic equation $a{{x}^{2}}+bx+c=0$.
Then, its roots are $\alpha $ and $\beta $. So, the sum of the roots and the product of the roots is formulated as:
$\alpha +\beta =\dfrac{-b}{a};\alpha \beta =\dfrac{c}{a}$
Complete step by step solution:The given quadratic equation is ${{x}^{2}}+px+q=0$ and its roots are $\alpha $ and $\beta $.
So, comparing the given equation with the standard equation, we get
$a=1;b=p;c=q$
Then, the sum of the roots of the given equation is obtained as
$\alpha +\beta =\dfrac{-p}{1}=-p\text{ }...(1)$
And the product of the roots of the given equation is
$\alpha \beta =\dfrac{q}{1}=q\text{ }...(2)$
It is given that the sum of the roots of the equation is 3 times their difference. Thus, we can write
$(\alpha +\beta )=3(\alpha -\beta )$
On simplifying, we get
$\begin{align}
& \Rightarrow \alpha +\beta =3(\alpha -\beta ) \\
& \Rightarrow \alpha +\beta =3\alpha -3\beta \\
& \Rightarrow 2\alpha =4\beta \\
& \Rightarrow \alpha =2\beta \text{ }...(3) \\
\end{align}$
On substituting (3) in (1), we get
$\begin{align}
& \alpha +\beta =-p \\
& \Rightarrow 2\beta +\beta =-p \\
& \Rightarrow 3\beta =-p \\
& \Rightarrow \beta =\dfrac{-p}{3}\text{ }...(4) \\
\end{align}$
On substituting (3) in (2), we get
$\begin{align}
& \alpha \beta =q \\
& \Rightarrow (2\beta )\beta =q \\
& \Rightarrow 2{{\beta }^{2}}=q\text{ }...(5) \\
\end{align}$
Then, substituting (4) in (5), we get
$\begin{align}
& 2{{\beta }^{2}}=q \\
& \Rightarrow 2{{\left( \dfrac{-p}{3} \right)}^{2}}=q \\
& \Rightarrow \dfrac{2{{p}^{2}}}{9}=q \\
& \therefore 2{{p}^{2}}=9q \\
\end{align}$
Thus, this is the required expression.
Option A is correct
Note: Here, we need to remember the sum and product formulae. So, we can substitute these values in the given relation to find an expression. And be sure with the signs in the formulae. Otherwise, the entire calculation may go in the wrong way.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges

NEET Total Marks 2025
