1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the
coefficients.
(i) ${x^2 – 2x – 8}$
Solution:
Factorise the polynomial by splitting the middle term $2x$ such that the product of the splits terms is equal
to the the product of the $x^2$ and constant term of the polynomial.
So, $-2x$ can be split as $-4x$ and $2x$ because $-4x^2$x $2x$ = $-8x^2$ and $-4x$ - $2x$ =$-2x$
Therefore, ${x^2 – 2x – 8}$ =$x^2 +2x - 4x - 8$
=$x(x+2) - 4(x+2)$
=$(x+2)(x-4)$
Therefore the zeroes of the polynomial is: $(x+2)=0$ and $(x-4)=0$
i.e $x= -2 $ and $x=4$
Since the relationship between zeroes and coefficients are given as:
Sum of Zeroes = $-{coefficient of x}/{coefficient of x^2}$ = $-b/a$
L.H.S = (-2)+(4)= 2
R.H.S = $-(-2)/(1)$ = 1.
Therefore L.H.S = R.H.S
Product of Zeroes = ${coefficient of constant term }/{coefficient of x^2}$
L.H.S = (-2)x(4) = -8
R.H.S = $-8/1$ = -8
Therefore L.H.S = R.H.S
(ii) ${4s^2 – 4s + 1}$
Solution:
${4s^2 – 4s + 1}$ = $4s^2-2s-2s + 1$
= $2s(2s-1)-1(2s-1)$
=$(2s-1)(2s-1)$
Therefore the zeroes of the polynomial are: (2s-1)=0 and (2s-1)=0
i.e $s=1/2$ and $s=1/2$
Since the relationship between zeroes and coefficients are given as:
Sum of Zeroes = $-{coefficient of x}/{coefficient of x^2}$ = $-b/a$
L.H.S = $1/2$ + $1/2$ =$1$
R.H.S = $-(-4)/4$ = $1$
Therefore L.H.S = R.H.S
Product of Zeroes = ${coefficient of constant term }/{coefficient of x^2}$ =$c/a$
L.H.S = $(1/2)×(1/2)$ = $1/4$
R.H.S = $1/4$
Therefore L.H.S = R.H.S
(iii) ${6x^2 – 3 – 7x}$
Solution:
Arrange the polynomial in order. i.e ${6x^2 – 3 – 7x}$ = $6x^2 - 7x - 3$
$6x^2 - 7x - 3$ = $6x^2-9x+2x-3$
= $3x(2x-3) +1(2x-3)$
=$(2x-3)(3x+1)$
Therefore the zeroes of the polynomial are: $(2x-3)=0$ and $(3x+1)=0$
i.e $x=3/2$ and $x= -1/3$
Since the relationship between zeroes and coefficients are given as:
Sum of Zeroes = $-{coefficient of x}/{coefficient of x^2}$ = $-b/a$
L.H.S= $3/2$ + $(-1/3)$ = $(9-2)/6$ =$7/6$
R.H.S = $-(-7)/6$ = $7/6$
Therefore L.H.S = R.H.S
Product of Zeroes = ${coefficient of constant term }/{coefficient of x^2}$ =$c/a$
L.H.S = $3/2$ × $-1/3$ = $-1/2$
R.H.S = $(-3)/6$ = $-1/2$
Therefore L.H.S = R.H.S
(iv) ${4u^2 + 8u}$
Solution:
${4u^2 + 8u}$ = $4u(u+2)$
Therefore the zeroes of the polynomial are: $4u=0$ and $u+2=0$
i.e $u=0$ and $u=-2$
Since the relationship between zeroes and coefficients are given as:
Sum of Zeroes = $-{coefficient of x}/{coefficient of x^2}$ = $-b/a$
L.H.S = $0+ (-2) = -2$
R.H.S = $-8/4$ = $-2$
Therefore L.H.S = R.H.S
Product of Zeroes = ${coefficient of constant term }/{coefficient of x^2}$ =$c/a$
L.H.S = $0 × ( -2)$ =$0$
R.H.S = $0/4$ = $0$
Therefore L.H.S =R.H.S
(v) ${t^2 – 15}$
Solution:
${t^2 – 15}$ = $t^2 - (√15)^2$
= $(t- √15)(t+√15)$
Therefore the zeroes of the polynomial are: $(t- √15)=0$ and $(t+√15)=0$
i.e $t=√15 $ and $t= -√15$
Since the relationship between zeroes and coefficients are given as:
Sum of Zeroes = $-{coefficient of x}/{coefficient of x^2}$ = $-b/a$
L.H.S = $ √15 + (-√15) = 0$
R.H.S = $0/1$ = $0$
Therefore L.H.S = R.H.S
Product of Zeroes = ${coefficient of constant term }/{coefficient of x^2}$ =$c/a$
L.H.S = $ √15 × (-√15)$ = $-15$
R.H.S = $(-15)/1$ = $-15$
Therefore L.H.S =R.H.S
(vi) ${3x^2 – x – 4}$
Solution:
${3x^2 – x – 4}$ = $3x^2 + 3x - 4x - 4$
= $3x(x+1) -4(x+1)$
=$(x+1)(3x-4)$
Therefore the zeroes of the polynomial are: $(x+1)=0 $ and $(3x-4)=0$
i.e $x=-1$ and $x=4/3$
Since the relationship between zeroes and coefficients are given as:
Sum of Zeroes = $-{coefficient of x}/{coefficient of x^2}$ = $-b/a$
L.H.S = $(-1)+4/3$ = $(-3+4)/3$ = $1/3$
R.H.S = $-(-1)/3$ = $1/3$
Therefore L.H.S =R.H.S
Product of Zeroes = ${coefficient of constant term }/{coefficient of x^2}$ =$c/a$
L.H.S = $(-1) × (4/3)$ = $-4/3$
R.H.S = $(-4)/3$ = $-4/3$
Therefore L.H.S =R.H.S
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) ${1/4,1}$
Solution:
If α and β are the zeroes of a polynomial,
then a polynomial is express as its sum and products of its zeroes as given below
$p(x)$= $kx^2 – k(α + β)x + k αβ$
$p(x)$ = $kx^2 -(1/4)kx$ +$(1)k$
if $k=1$, then
$p(x)$ =$4x^2 -x +4$
(ii) ${√2 , 1/3}$
Solution:
If α and β are the zeroes of a polynomial,
then a polynomial is express as its sum and products of its zeroes as given below
$p(x)$= $kx^2 – k(α + β)x + k αβ$
$p(x)$=$kx^2 - k(√2)x +k (1/3)$
$p(x)$ =$3x^2 -3√2x +1 $
(iii) ${0,√5 }$
Solution:
If α and β are the zeroes of a polynomial,
then a polynomial is express as its sum and products of its zeroes as given below
$p(x)$= $kx^2 – k(α + β)x + k αβ$
= $x^2 -(0)x + √5 $
=$ x^2 + √5$
(iv) ${1,1}$
Solution:
If α and β are the zeroes of a polynomial,
then a polynomial is express as its sum and products of its zeroes as given below
$p(x)$= $kx^2 – k(α + β)x + k αβ$
= $x^2 – (1)x + 1$
$p(x)$ =$x^2 - x + 1$
(v) ${-1/4,1/4}$
Solution:
If α and β are the zeroes of a polynomial,
then a polynomial is express as its sum and products of its zeroes as given below
$p(x)$= $kx^2 – k(α + β)x + k αβ$
= $x^2 – (-1/4)x + (1/4)$
=$4x^2 + x + 1$
(vi) ${4,1}$
Solution:
If α and β are the zeroes of a polynomial,
then a polynomial is express as its sum and products of its zeroes as given below
$p(x)$= $kx^2 – k(α + β)x + k αβ$
= $x^2 – (4)x + 1$
= $x^2 – 4x + 1$
