
If the ratio of the roots of $a{x^2} + bx + c = 0$ is same as the roots of $p{x^2} + qx + r = 0$ then:
A. $\dfrac{b}{{ac}} = \dfrac{q}{{pr}}$
B. $\dfrac{{{b^2}}}{{ac}} = \dfrac{{{q^2}}}{{pr}}$
C. $\dfrac{{2b}}{{ac}} = \dfrac{{{q^2}}}{{pr}}$
D. None of these
Answer
162.6k+ views
Hint: In this question, we are given two quadratic equation i.e., $a{x^2} + bx + c = 0$ and $p{x^2} + qx + r = 0$. We have to find the ratio of their roots where the ratio of the roots of the equations is equal. Firstly, calculate the sum and the product of roots. Then, write the ratio of roots in two ways and add them together.
Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$
Complete step by step Solution:
Given that,
There are two quadratic equations $a{x^2} + bx + c = 0$ and $p{x^2} + qx + r = 0$.
Let, \[\alpha ,\beta \] are the roots of $a{x^2} + bx + c = 0$
And \[\gamma ,\delta \] are the roots of $p{x^2} + qx + r = 0$
Now, using the formula of sum and product of the roots in the equations
We get,
For $a{x^2} + bx + c = 0$,
Sum, $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$ -------(1)
For $p{x^2} + qx + r = 0$.
Sum, $\gamma + \delta = \dfrac{{ - q}}{p}$ and $\gamma \delta = \dfrac{r}{p}$ -------(2)
According to the question,
The ratio of the root is equal. It implies that,
\[\dfrac{\alpha }{\beta } = \dfrac{\gamma }{\delta }\] ---------(3)
And \[\dfrac{\beta }{\alpha } = \dfrac{\delta }{\gamma }\] ---------(4)
Adding equations (3) and (4),
\[\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{\gamma }{\delta } + \dfrac{\delta }{\gamma }\]
On simplifying we get,
\[\dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} = \dfrac{{{\gamma ^2} + {\delta ^2}}}{{\gamma \delta }}\]
Adding $2$ on both sides,
\[\dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} + 2 = \dfrac{{{\gamma ^2} + {\delta ^2}}}{{\gamma \delta }} + 2\]
\[\dfrac{{{\alpha ^2} + {\beta ^2} + 2\alpha \beta }}{{\alpha \beta }} = \dfrac{{{\gamma ^2} + {\delta ^2} + 2\gamma \delta }}{{\gamma \delta }}\]
(Converting the numerator using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$)
\[\dfrac{{{{\left( {\alpha + \beta } \right)}^2}}}{{\alpha \beta }} = \dfrac{{{{\left( {\gamma + \delta } \right)}^2}}}{{\gamma \delta }}\]
Substituting the values from equations (1) and (2),
\[\dfrac{{{{\left( {\dfrac{{ - b}}{a}} \right)}^2}}}{{\left( {\dfrac{c}{a}} \right)}} = \dfrac{{{{\left( {\dfrac{{ - q}}{p}} \right)}^2}}}{{\left( {\dfrac{r}{p}} \right)}}\]
On solving it further,
\[\dfrac{{{b^2}}}{{ac}} = \dfrac{{{q^2}}}{{pr}}\]
Hence, the correct option is (B).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$
Complete step by step Solution:
Given that,
There are two quadratic equations $a{x^2} + bx + c = 0$ and $p{x^2} + qx + r = 0$.
Let, \[\alpha ,\beta \] are the roots of $a{x^2} + bx + c = 0$
And \[\gamma ,\delta \] are the roots of $p{x^2} + qx + r = 0$
Now, using the formula of sum and product of the roots in the equations
We get,
For $a{x^2} + bx + c = 0$,
Sum, $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$ -------(1)
For $p{x^2} + qx + r = 0$.
Sum, $\gamma + \delta = \dfrac{{ - q}}{p}$ and $\gamma \delta = \dfrac{r}{p}$ -------(2)
According to the question,
The ratio of the root is equal. It implies that,
\[\dfrac{\alpha }{\beta } = \dfrac{\gamma }{\delta }\] ---------(3)
And \[\dfrac{\beta }{\alpha } = \dfrac{\delta }{\gamma }\] ---------(4)
Adding equations (3) and (4),
\[\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{\gamma }{\delta } + \dfrac{\delta }{\gamma }\]
On simplifying we get,
\[\dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} = \dfrac{{{\gamma ^2} + {\delta ^2}}}{{\gamma \delta }}\]
Adding $2$ on both sides,
\[\dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} + 2 = \dfrac{{{\gamma ^2} + {\delta ^2}}}{{\gamma \delta }} + 2\]
\[\dfrac{{{\alpha ^2} + {\beta ^2} + 2\alpha \beta }}{{\alpha \beta }} = \dfrac{{{\gamma ^2} + {\delta ^2} + 2\gamma \delta }}{{\gamma \delta }}\]
(Converting the numerator using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$)
\[\dfrac{{{{\left( {\alpha + \beta } \right)}^2}}}{{\alpha \beta }} = \dfrac{{{{\left( {\gamma + \delta } \right)}^2}}}{{\gamma \delta }}\]
Substituting the values from equations (1) and (2),
\[\dfrac{{{{\left( {\dfrac{{ - b}}{a}} \right)}^2}}}{{\left( {\dfrac{c}{a}} \right)}} = \dfrac{{{{\left( {\dfrac{{ - q}}{p}} \right)}^2}}}{{\left( {\dfrac{r}{p}} \right)}}\]
On solving it further,
\[\dfrac{{{b^2}}}{{ac}} = \dfrac{{{q^2}}}{{pr}}\]
Hence, the correct option is (B).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
