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Relationship of Zeroes With Coefficients of Quadratic Polynomial

Polynomial - Concepts

Polynomial Expression

Degree of a Term

Degree of a Polynomial

Standard Form For a Polynomial

Multiplication of Polynomials

Factor And Remainder Theorem

Relationship of Zeroes With Coefficients of Quadratic Polynomial

Graphical Representation of A Polynomial

Forming a Quadratic Polynomial When Zeroes are Given

Division of Polynomial by another Polynomial Long Division

Factorization of Cubic Polynomial

Relationship of Zeros With Coefficients of Cubic Polynomial

Class - CAT Subjects

Concept Explanation

Relationship of Zeroes With Coefficients of Quadratic Polynomial

Relationship of Zeroes With Coefficients Of Quadratic Polynomial:

If $p(x)ax+b,aneq 0,;and;alpha$ (read as alpha) is a zero of p(x) then

$alpha =-frac{b}{a}=frac{constant;term}{coefficient;of;x}$

If $p(x)=ax^{2}+bx+c, a neq 0$ and $alpha ,beta$ ( read as beta) are two zeroes of p(x) then

$alpha + beta =frac{-b}{a}=-frac{coefficient;of;x}{coefficient;of;x^{2}}$

$alphabeta =frac{c}{a}=frac{constant;term}{coefficient;of;x^{2}}$

We can use these reltaions to obtain values of several symmetric expressions involving $alpha$ and $beta$ in terms of a, b and c.

An expressons in $alpha$ and $beta$ is said to be symmetric if it remaiins unchanged when $alpha$ and $beta$ are intercharged. For instance, $alpha ^{2}+beta ^{2},alpha ^{2}-alpha beta + beta ^{2}$ and $frac{(alpha ^{2}+beta ^{2})}{alpha beta }$ are symmetric expressions in $alpha$ and $beta$ .

Example: If a and c are such that the quadratic polynomial $ax^{2}-5x+c$ has 10 as the sum of zeroes and also as the product of zeroes, fiind a and c.

Solution: Let $alpha ,beta$ be the zeroes of $ax^{2}-5x+c.$ Then

$alpha +beta =frac{5}{a},alpha beta =frac{c}{a}$

As $10=alpha +beta =alpha beta ,$ we get

$10=frac{5}{a},10=frac{c}{a}$

$Rightarrow$ $a=frac{5}{10}=frac{1}{2}$ and c = 10a = 5

Thus, $a=frac{1}{2},c=5$

Example : If $alpha$ and $beta$ are zeros of the quadratic polynomial $x^{2}-5x+6$ . find the value of $alpha ^{4}+beta ^{4}.$

Solution: $alpha +beta =5,alpha beta =6$

We first find $alpha ^{2}+beta ^{2}.$ We have

$alpha ^{2}+beta ^{2}=(alpha +beta )^{2}-2alpha beta =25-2(6)=13$

$alpha ^{4}+beta ^{4}=(alpha ^{2}+beta ^{2})^{2}-2alpha ^{2}beta ^{2}$

$=13^{2}-2(6)^{2}$

$=169-72=97$

Thus, $alpha ^{4}+beta ^{4}=97$

Example: If $large alpha ,beta$ are zeros of $large p(x)=ax^{2}+bx+c$ then obtain quadratic polynomials whose zeros are $large -alpha ,-beta$

Solution: As $large alpha ,beta$ are zeros of $large p(x)=ax^{2}+bx+c$

$large alpha +beta =-frac{b}{a},alpha beta =frac{c}{a}$

$large S=(-alpha )+(-beta )=-(alpha +beta )=frac{b}{a}$

$large P=(-alpha )(-beta )=alpha beta =frac{c}{a}$

Thus, the quadratic expression whose zeros are $large -alpha ,-beta$ is

$large x^{2}-Sx+P$

or $large x^{2}-frac{b}{a}x+frac{c}{a}$

or $large ax^{2}-bx+c$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

For which value of 'a' will the polynomial $large (x-4)^2+(2a-373)^2$ have equal roots.
A. 373		B. $large frac{373}{2}$
C. 0		D. $large -frac{373}{2}$
Right Option : B
View Explanation