
If $\alpha $ and $\beta$ are roots of the equation $ax^{2}+2bx+c=0$, then
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}$ is equal to
A. $\dfrac{2b}{c}$
B.$\dfrac{2b}{\sqrt{ac}}$
C. $\dfrac{-2b}{\sqrt{ac}}$
D. $\dfrac{-b}{\sqrt{ac}}$
Answer
163.5k+ views
Hint:The values of the variable that satisfy a quadratic equation are known as its roots. The "solutions" or "zeros" of the quadratic equation are other names for them. We need to first simplify the given equation whose value we need to find and then substitute the value of the sum of the roots and the product of the roots in the simplified equation.
Formula Used:
Sum of roots,
$\alpha +\beta =\dfrac{-b}{a}$
Product of the roots,
$\alpha \beta =\dfrac{c}{a}$
Complete step by step Solution:
The given equation is
$ax^{2}+2bx+c=0$ which is quadratic in nature.
Let the roots of the quadratic equation be $\alpha$and $\beta$
Now, substituting the values of respective coefficients in the formula of the sum of roots and product of roots, we get
$\alpha +\beta =\dfrac{-b}{a}=\dfrac{-2b}{a}$
Also,
$\alpha \beta =\dfrac{c}{a}$
We need to find
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}$
By taking lcm of the above equation
$\dfrac{\left ( \sqrt{\alpha ^{2}} \right )+\left ( \sqrt{\beta^{2}} \right )}{\sqrt{\alpha \beta }}$
$= \dfrac{\alpha +\beta }{\sqrt{\alpha \beta }}$
$=\dfrac{\dfrac{-2b}{a}}{\sqrt{\dfrac{c}{a}}}$
Therefore,
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}=\dfrac{-2b}{\sqrt{ac}}$
Hence, the correct option is (C).
Additional Information:We know that the sum of the roots of a quadratic equation is the negative coefficient of the second term of the generalized form which is linear in nature divided by the coefficient of the first term of the generalized form of the equation which is quadratic in nature.
Also, the product of the roots is the third term of the generalized form of the equation which is constant in nature divided by the coefficient of the first term of the generalized form of the
equation which is quadratic in nature.
Note: Students must make sure that the simplified version of the given equation is either in the form sum of roots or the product of roots or a combination of both of them. This needs to be done so that we have values to substitute which will give our ultimate result.
Formula Used:
Sum of roots,
$\alpha +\beta =\dfrac{-b}{a}$
Product of the roots,
$\alpha \beta =\dfrac{c}{a}$
Complete step by step Solution:
The given equation is
$ax^{2}+2bx+c=0$ which is quadratic in nature.
Let the roots of the quadratic equation be $\alpha$and $\beta$
Now, substituting the values of respective coefficients in the formula of the sum of roots and product of roots, we get
$\alpha +\beta =\dfrac{-b}{a}=\dfrac{-2b}{a}$
Also,
$\alpha \beta =\dfrac{c}{a}$
We need to find
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}$
By taking lcm of the above equation
$\dfrac{\left ( \sqrt{\alpha ^{2}} \right )+\left ( \sqrt{\beta^{2}} \right )}{\sqrt{\alpha \beta }}$
$= \dfrac{\alpha +\beta }{\sqrt{\alpha \beta }}$
$=\dfrac{\dfrac{-2b}{a}}{\sqrt{\dfrac{c}{a}}}$
Therefore,
$\sqrt{\dfrac{\alpha }{\beta }}+\sqrt{\dfrac{\beta }{\alpha }}=\dfrac{-2b}{\sqrt{ac}}$
Hence, the correct option is (C).
Additional Information:We know that the sum of the roots of a quadratic equation is the negative coefficient of the second term of the generalized form which is linear in nature divided by the coefficient of the first term of the generalized form of the equation which is quadratic in nature.
Also, the product of the roots is the third term of the generalized form of the equation which is constant in nature divided by the coefficient of the first term of the generalized form of the
equation which is quadratic in nature.
Note: Students must make sure that the simplified version of the given equation is either in the form sum of roots or the product of roots or a combination of both of them. This needs to be done so that we have values to substitute which will give our ultimate result.
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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
