
The quadratic equation with rational, coefficient the sum of the squares of whose root is 40 and the sum of the cubes of whose root is 208 is
A. \[{{x}^{2}}+4x+12=0\]
B. \[{{x}^{2}}-4x-12=0\]
C. \[{{x}^{2}}-4x+12=0\]
D. \[{{x}^{2}}+4x-12=0\]
Answer
164.4k+ views
Hint: In this question, we have given four equations. We have to find an equation which satisfies the conditions given in the equation. To find out that equation, we make the equations from the conditions given in the question and by solving those equations with the help of formulas, we get our desirable answer.
Complete step by step solution:
Given are the four quadratic equations.
We have to find out the equation whose sum of the square of root is 40 and sum of the cubes of root is 208.
Let $\alpha ,\beta $be its roots.
That means ${{\alpha }^{2}}+{{\beta }^{2}}=40$ and ${{\alpha }^{3}}+{{\beta }^{3}}=208$ (given in the equation)
Now suppose $\alpha +\beta =m,\alpha \beta =n$
${{\alpha }^{2}}+{{\beta }^{2}}=40$
And we know formula of $({{x}^{3}}+{{y}^{3}})=(x+y)({{x}^{2}}-xy+{{y}^{2}})$
By putting x = $\alpha $and y = $\beta $ in the above equation, we get
$({{\alpha }^{3}}+{{\beta }^{3}})=(\alpha +\beta )({{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}})$ ${{x}^{2}}-4x-12=0$
$(\alpha +\beta )({{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}})=208$
Now put $(\alpha +\beta )=m$and $\alpha \beta =n$in the above equation, we get
$m({{\alpha }^{2}}+{{\beta }^{2}}-n)=208$
As given in the question ${{\alpha }^{2}}+{{\beta }^{2}}=40$, we put this value in the above equation, we get
$m(40-n)=208$ . . . . . . . .. . . . . . . . . . . . . . . . .(1)
We know ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
By putting $a=\alpha $ and $b=\beta $, we get
${{(\alpha +\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta $
By putting value of m and n in the above equation, we get
${{m}^{2}}=40+2n$ ( as ${{\alpha }^{2}}+{{\beta }^{2}}=40$)
From this, we get $n=\frac{{{m}^{2}}-40}{2}$
Put this value of n in equation (1) , we get
$m\left[ \left( 40 \right)-\left\{ \frac{{{m}^{2}}-40}{2} \right\} \right]=208$
By solving this equation, we get
${{m}^{3}}-120m+216=0$
By using long division to solve the above equation, we get
$(m-4)({{m}^{2}}+4m-104)=0$
By solving, we get (m-4) =0
$m=4$
Negative value is not possible.
So $m=4,n=-12$
From there, we get a quadratic equation ${{x}^{2}}-4x-12=0$
Option (B) is correct.
Note: To solve these types of questions, we must know from which method we are able to get the answer easily. Sometimes we get answers from the options easily. We have to take care while putting the formulas in the equation then we get our answer.
Complete step by step solution:
Given are the four quadratic equations.
We have to find out the equation whose sum of the square of root is 40 and sum of the cubes of root is 208.
Let $\alpha ,\beta $be its roots.
That means ${{\alpha }^{2}}+{{\beta }^{2}}=40$ and ${{\alpha }^{3}}+{{\beta }^{3}}=208$ (given in the equation)
Now suppose $\alpha +\beta =m,\alpha \beta =n$
${{\alpha }^{2}}+{{\beta }^{2}}=40$
And we know formula of $({{x}^{3}}+{{y}^{3}})=(x+y)({{x}^{2}}-xy+{{y}^{2}})$
By putting x = $\alpha $and y = $\beta $ in the above equation, we get
$({{\alpha }^{3}}+{{\beta }^{3}})=(\alpha +\beta )({{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}})$ ${{x}^{2}}-4x-12=0$
$(\alpha +\beta )({{\alpha }^{2}}-\alpha \beta +{{\beta }^{2}})=208$
Now put $(\alpha +\beta )=m$and $\alpha \beta =n$in the above equation, we get
$m({{\alpha }^{2}}+{{\beta }^{2}}-n)=208$
As given in the question ${{\alpha }^{2}}+{{\beta }^{2}}=40$, we put this value in the above equation, we get
$m(40-n)=208$ . . . . . . . .. . . . . . . . . . . . . . . . .(1)
We know ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
By putting $a=\alpha $ and $b=\beta $, we get
${{(\alpha +\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta $
By putting value of m and n in the above equation, we get
${{m}^{2}}=40+2n$ ( as ${{\alpha }^{2}}+{{\beta }^{2}}=40$)
From this, we get $n=\frac{{{m}^{2}}-40}{2}$
Put this value of n in equation (1) , we get
$m\left[ \left( 40 \right)-\left\{ \frac{{{m}^{2}}-40}{2} \right\} \right]=208$
By solving this equation, we get
${{m}^{3}}-120m+216=0$
By using long division to solve the above equation, we get
$(m-4)({{m}^{2}}+4m-104)=0$
By solving, we get (m-4) =0
$m=4$
Negative value is not possible.
So $m=4,n=-12$
From there, we get a quadratic equation ${{x}^{2}}-4x-12=0$
Option (B) is correct.
Note: To solve these types of questions, we must know from which method we are able to get the answer easily. Sometimes we get answers from the options easily. We have to take care while putting the formulas in the equation then we get our answer.
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
