If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

Asked by Last Modified  

5 Answers

Learn Mathematics +1 Class IX-X Tuition

Follow 4
Answer

Please enter your answer

Aditya Kumar

Mathematics

zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since...
read more
zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since f(x)= x^3 - ax^2 +bx - c f(A)= A^3 - aA^2 +bA -c put A= a/3 (a/3)^3 -a(a/3)^2 +b(a/3) -c=0 (a^3)/27 - (a^3)/9 +ab/3 -c = 0 (a^3) -3(a^3) +9ab -27c =0 -2(a^3) +9ab - 27c = 0 2a^3 -9ab +27c = 0 hence PROVED read less
Comments
Dislike Bookmark

let l , m , n are zeroes of given polynomial. now sum of l+m+n = a product lmn = c product of zeros taken two at time lm+mn+nl = b also it is given that zeros of polynomial are in A. P . then 2m= l+ n from these equations we get a = 3m , b=2m2 + ln. now putting values...
read more
let l , m , n are zeroes of given polynomial. now sum of l+m+n = a product lmn = c product of zeros taken two at time lm+mn+nl = b also it is given that zeros of polynomial are in A. P . then 2m= l+ n from these equations we get a = 3m , b=2m2 + ln. now putting values of a, b , c in LHL we get 0. read less
Comments
Dislike Bookmark

Tutor

Let p-1, p, p+1 be the zeroes of the given polynomial. p-1+p+p+1=a 3p=a (p-1)p+p(p+1)+(p-1)(p+1)=b p2-p+p2+p+p2-1=b 3p2-1=b (p-1)p(p+1)=c (p2-1)p=c p3-p=c Consider 2a3-9ab+27c 2(3p)3-9(3p)(3p2-1)+27(p3-p)= 54p3-81p3+27p+27p3-27p= 81p3-81p3=0 Therefore 2a3-9ab+27c=0
Comments
Dislike Bookmark

Trainer

Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily,...
read more
Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily, we get a new equation by putting the (m+d), =>m^3+d^3+3m*d^2-a*m^2-a*d*d^2-2a*m*d+b*m+b*d-c=0---------------------(ii) Again, Putting the value of m in polynomial, we have m^3-a*m^2+b*m-c=0-----------------------------------------(iii) Adding equation (i) & (ii), we have=>2*m^3+6*m*d^2-2a*m^2-2a*d^2+2b*m-2*c=0 =>2*(3^3-a*m^2+b*m-c)+6m*d^2-2*a*d^2=0 =>6*m*d^2=2*a*d^2 =>m=a/3------------------------------(iv) Now, Putting the value of m in the polynomial, =>(a/3)^3-a*(a/3)^2+b*a/3-c=0 =>a^3/27-a^3/9+a*b/3-c=0 =>a^3-3*a^3+9*a*b-27*c=0 =>2*a^3-9*a*b+27*c=0 (Proved) read less
Comments
Dislike Bookmark

As we know, sum of roots = -(coeff. of x2)/(coeff. of x3) = a Since the roots ( or zeros) of polynomial are in AP, then we can assume the roots to be A-D, A, A+D ( with D as common diff.) sum of roots = A-D+A+A+D = 3A =a therefore A = a/3 Since, A is a zero of the polynomial, then it should satisfy...
read more
As we know, sum of roots = -(coeff. of x2)/(coeff. of x3) = a Since the roots ( or zeros) of polynomial are in AP, then we can assume the roots to be A-D, A, A+D ( with D as common diff.) sum of roots = A-D+A+A+D = 3A =a therefore A = a/3 Since, A is a zero of the polynomial, then it should satisfy the eqn. put A=a/3 in eqn (a^3)/27 - (a^3)/9 + (a*b)/3 - c = 0 Hence, 2a3-9ab+27c = 0 read less
Comments
Dislike Bookmark

View 3 more Answers

Related Questions

What is algebraic geometry?
Algebraic geometry is a branch of mathematics that studies the solution sets of polynomial equations. It uses tools from both algebra and geometry to study geometric objects, such as curves and surfaces, that can be defined by polynomial equations.
Franziska
0 0
7
If sin A=1/2, then find the value of cos A.
sin (A/2)= 1/2 =sin (A/2)=sin (30°)=sin(180-30) A/2= 30° or 150° A= 60° or 300° sin(A)=sin (60°)=√3/2 sin(A)=sin (300°)=sin(360°+ -60°)=sin(-60°)=-sin (60°)=(-√3)/2 sin...
Malvika
Suppose that I am trying to borrow money from you to finance my business, and I promise to repay you Rs. 1,000 quarterly for two years. If your opportunity cost of funds is 10%, how much are you willing to lend me?
Rs. 3636.36.
Soumya
0 0
5
Is it necessary to pay for computer tuition in class 10 ICSE?
YES
Ushashree
0 0
5
How long should a student of class 9 practice math?
For an average and a slow learner 2 hours per day is required whereas for a above average just enough to have at glance. It depends on the strengths of the child on the subject.
Bhushan
0 0
5

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons

Class-10 L-1 Real Numbers Ex-1.1 Q4

Ashish Bhatia

0 0
0
Class-10 L-5 Arithmetic Progressions Ex-5.2 Q17

Ashish Bhatia

0 0
0
Class-10 L-4 Quadratic Equations Ex-4.3 Q1 (iii)

Ashish Bhatia

0 0
0
Class-10 L-1 Real Numbers Ex-1.1 Q7

Ashish Bhatia

0 0
0
PYQ MATHS GRADE 10 Ways to Extract questions
How to Find Important Questions for CBSE Exams: 1. Go to Google and type “PYQ Important Questions Learn CBSE”. 2. To get topic-specific questions, type “PYQ Grade 10 Mathematics ”. 3....

Siddharth

0 0
0

Recommended Articles

Meet Sandhya R, a B.Sc tutor from Bangalore

Sandhya is a proactive educationalist. She conducts classes for CBSE, PUC, ICSE, I.B. and IGCSE. Having a 6-year experience in teaching, she connects with her students and provides tutoring as per their understanding. She mentors her students personally and strives them to achieve their goals with ease. Being an enthusiastic...

Read full article >

Quest Academy - Institute of the month

Quest Academy is a professional Bangalore based NEET and JEE (Main + Advanced) training institute. The academy was incorporated in 2015 to cater to the needs of students, who aim to crack competitive exams by connecting with the best brains around. The institute helps students enhance their skills and capabilities through...

Read full article >

Meet Mohammad Wazid, a skilled trainer for...

Mohammad Wazid is a certified professional tutor for class 11 students. He has 6 years of teaching experience which he couples with an energetic attitude and a vision of making any subject easy for the students. Over the years he has developed skills with a capability of understanding the requirements of the students. This...

Read full article >

Meet Raghunandan.G.H, a B. Tech Tutor from...

Raghunandan is a passionate teacher with a decade of teaching experience. Being a skilled trainer with extensive knowledge, he provides high-quality BTech, Class 10 and Class 12 tuition classes. His methods of teaching with real-time examples makes difficult topics simple to understand. He explains every concept in-detail...

Read full article >

Find Class X Tuition near you
Class X Tuition in Delhi
Class X Tuition in Bangalore
Class X Tuition in Hyderabad
Class X Tuition in Kolkata
Class X Tuition in Mumbai
Class X Tuition in Chennai
Class X Tuition in Pune
Class X Tuition in Noida
Class X Tuition in Lucknow
Class X Tuition in Gurgaon
Online Tuition Class 10

Looking for Class 10 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you