# Exercise - 2.1 (Mathematics NCERT Class 10th)

Q.1     The graphs of y = p(x) are given in figures below for some polynomials p(x). Find the number of zeroes of p(x) , in each case.
(i)
(ii)
(iii)
(iv)
(v)

(vi)

Sol.
(i) There are no zeroes as the graph does not intersect the x-axis.
(ii) The number of zeroes is one as the graph intersects the x-axis at one point only.
(iii) The number of zeroes is three as the graph intersects the x-axis at three points.
(iv) The number of zeroes is two as the graph intersects the x-axis at two points.
(v) The number of zeroes is four as the graph intersects the x-axis at four points.
(vi) The number of zeroes is three as the graph intersects the x-axis at three points.

# Polynomials : Exercise - 2.2 (Mathematics NCERT Class 10th)

Q.1       Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :
(i) x22x8
(ii) 4s24s+1
(iii) 6x237x
(iv) 4u2+8u
(v) t215
(vi) 3x2x4

Sol.       (i) We have, x22x8 =x2+2x4x8
=x(x+2)4(x+2)
=(x+2)(x4)

The value of x22x8 is zero when the value of (x + 2) (x – 4) is zero, i.e.,
when x + 2 = 0 or x – 4 = 0 , i.e., when x = – 2 or x = 4.

So, The zeroes of x22x8 are – 2 and 4.
Therefore , sum of the zeroes = (– 2) + 4 = 2
=CoefficientofxCoefficientofx2
and product of zeroes = (– 2) (4) = – 8 =81
=ConstanttermCoefficientofx2

(ii) We have, 4s24s+1 =4s22s2s+1
=2s(2s1)1(2s1)
=(2s1)(2s1)
The value of 4s24s+1 is zero when the value of
(2s – 1) (2s – 1) is zero, i.e., when 2s – 1 = 0 or 2s – 1 = 0,
i.e., when s=12ors=12.
So, The zeroes of 4s24s+1are12and12
Therefore, sum of the zeroes =12+12=1
=CoefficientofsCoefficientofs2
and product of zeroes =(12)(12)=14
=ConstanttermCoefficientofs2
(iii) We have, 6x237x = 6x27x3
=6x29x+2x3
=3x(2x3)+1(2x3)
=(3x+1)(2x3)
The value of 6x237x is zero when the value of (3x + 1) (2x – 3) is zero, i.e., when 3x + 1 = 0 or 2x – 3 = 0, i.e, when x=13orx=32
So, The zeroes of 6x237xare13and32
Therefore, sum of the zeroes =13+32=76
=CoefficientofxCoefficientofx2
and product of zeroes =(13)(32)=12
=ConstanttermCoefficientofx2
(iv)  We have, 4u2+8u = 4u (u + 2)
The value of 4u2+8u is zero when the value of 4u(u + 2) is zero, i.e., when u = 0 or u + 2 = 0, i.e., when u = 0 or u = – 2.
So, The zeroes of 4u2+8u and 0 and – 2
Therefore, sum of the zeroes = 0 + (– 2) = – 2
=CoefficentofuCoefficientofu2
and , product of zeroes = (0) (–2) = 0
=ConstanttermCoefficientofu2
(v) We have t215 =(t15)(t+15)
The value of t215 is zero when the value of (t15)(t+15) is zero,
i.e., when
t15=0ort+15 = 0 i.e., when t=15ort=15
So, The zeroes of t215are15and15
Therefore , sum of the zeroes = 15+(15)=0
=CoefficientoftCoefficientoft2
and, product of the zeroes = (15)(15)=15
=ConstanttermCoefficientoft2
(vi) We have, 3x2x4 =  3x2+3x4x4
=3x(x+1)4(x+1)
=(x+1)(3x4)
The value of 3x2x4 is zero when the value of (x + 1) (3x – 4) is zero, i.e., when x + 1 = 0 or 3x – 4 = 0, i.e., when x = – 1 or  x=43.
So, The zeroes of 3x2x4are1and43
Therefore , sum of the zeroes =1+43=3+43
=13=CoefficientofxCoefficientofx2
and, product of the zeroes =(1)(43)=43
=ConstanttermCoefficientofx2

Q.2       Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 14,1
(ii) 2,13
(iii) 0,5
(iv) 1, 1
(v) 14,14

(vi) 4, 1
Sol.      (i) Let the polynomial be ax2+bx+c, and its zeroes be αandβ. Then ,
α+β=14=ba
and,                            αβ=1=44=ca
If a = 4,  then  b = – 1 and c = – 4.
Therefore, one quadratic polynomial which fits the given conditions is 4x2x4.

(ii) Let the polynomial be ax2+bx+c, and its zeroes be αandβ. Then,
α+β=2=323=ba
and                            αβ=13=ca
If a = 3, then b =32andc=1
So, One quadratic polynomial which fits the given conditions is 3x232x+1.

(iii) Let the polynomial  be ax2+bx+c, and its zeroes be αandβ. Then,

α+β=0=01=ba
and                           αβ=5=51=ca
If  a = 1, then b = 0 and c = 5
So, one quadratic polynomial which fits the given conditions is x20.x+5,i.e.,x2+5.

(iv) Let the polynomial be ax2+bx+c and its zeroes be αandβ. Then,
α+β=1
=(1)1=ba
a
nd                               αβ=1 =1