10th Maths 2.2
NCERT Class 10th solution of Exercise 2.1
Exercise 2.2
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
i) x2-2x-8 ii) 4s2-4s+1 iii) 6x2-3-7x
iv) 4u2+8u v) t2-15 vi) 3x2-x-4
Sol. :
i) Given quadratic equation x2-2x-8 compare with ax2+bx+c=0
where a=1,b=-2 and c=-8
Let
x2-2x-8=0
x2-4x+2x-8=0
x(x-4)+2(x-4)=0
(x-4)(x+2)=0
x-4=0
x=4
x+2=0
x=-2
Now, the sum of the zeroes=-ba
α+β=-(-2)1
4-2=-(-2)1
2=2
and the product of the zeroes =ca
α×β=-81
4×(-2)=-8
-8=-8
Answer :
The required zeroes 4 or -2 and the relation of zeroes & coefficient are verified.
ii) Given quadratic equation 4s2-4s+1 compare with ax2+bx+c=0
where a=4,b=-4 and c=-1
Let
4s2-4s+1=0
4s2-2s-2s+1=0
2s(2s-1)-1(2s-1)=0
(2s-1)(2s-1)=0
2s-1=0
s=12
2s-1=0
s=12
Now, the sum of the zeroes =-ba
α+β=-(-44)
12+12=1
1=1
and the product of the zeroesca
α×β=14
12×12=14
14=14
Answer :
The required zeroes 14,14 and the relation of zeroes & coefficient are verified.
iii) Given quadratic equation 6x2-3-7x=6x2-7x-3 compare with ax2+bx+c=0
where a=6,b=-7 and c=-3
Let
6x2-7x-3=0
6x2-9x+2x-3=0
3x(2x-3)+1(2x-3)=0
(2x-3)(3x+1)=0
2x-3=0
x=32
3x+1=0
x=-13
Now, the sum of the zeroes=-ba
α+β=-(-76)
32+(-13)=76
76=76
and the product of the zeroes=ca
α×β=-36
32×(-13)=-12
-12=-12
Answer :
The required zeroes 23,-13 and the relation of zeroes & coefficient are verified.
iv) Given quadratic equation 4u2+8u compare with ax2+bx+c=0
where a=4,b=8 and c=0
Let
4u2+8u=0
4u(u+2)=0
4u=0
u=0
u+2=0
u=-2
Now, sum of the zeroes=-ba
α+β=-84
0+(-2)=
-2 = -2
and product of the zeroes=c/a
alpha times beta=0/4
0times-2 = 0
0=0
Answer :
The required zeroes are 0, -2, and the relation of zeroes & coefficient are verified.
v) Given quadratic equation t^2-15 compare with ax^2+bx+c=0
where a=1, b=0 and c=-15
Let
t^2-15=0
t^2-(sqrt15)^2=0
(t-sqrt15)(t+sqrt15)=0
t-sqrt15=0
t=sqrt15
t+sqrt15=0
t=-sqrt15
Now, the sum of the zeroes -b/a
alpha+beta=-(0/(-sqrt15))
sqrt15+(-sqrt15)=0
0=0
and the product of the zeroes = c/a
alpha times beta=-15
sqrt15 times (-sqrt15) = -15
-15 = -15
Answer :
The required zeroes are sqrt15, -sqrt15 and the relation of zeroes & coefficient are verified.
vi) Given quadratic equation 3x^2-x-4 compare with ax^2+bx+c=0
where a=3, b=-1 and c=-4
Let
3x^2-x-4=0
3x^2-4x+3x-4=0
x(3x-4)+1(3x-4)=0
(3x-4)(x+1)=0
3x-4=0
x=4/3
x+1=0
x=-1
Now, sum of the zeroes= -b/a
alpha+beta= - (-1/3)
4/3+(-1)=1/3
1/3=1/3
and product of the zeroes=c/a
alpha times beta= -4/3
4/3times (-1) = -4/3
-4/3 = -4/3
Answer :
The required zeroes are 4/3, -1, and the relation of zeroes & coefficient are verified.
Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
i) 1/4, -1 ii) sqrt2, 1/3 iii) 0, sqrt5
iv) 1, 1 v) -1/4, 1/4 vi) 4, 1
Sol. :
i) Let the required quadratic polynomial is p(x)=ax^2+bx+c and its zeroes are alpha and beta. Then we have (according to question)
alpha+beta=-b/a=1/4=-(-1/4)
alpha times beta=c/a=(-4)/4=-1
a=4, b=-1, and c=-4
p(x)=4x^2-x-4
Answer :
Thus, the required quadratic polynomial is 4x^2-x-4.
ii) Let the required quadratic polynomial is p(x)=ax^2+bx+c and its zeroes are alpha and beta. Then we have (according to question)
alpha+beta=-b/a=sqrt2=-((-3sqrt2)/3)
alpha times beta=c/a=1/3
a=3, b=-3sqrt2, and c=1
p(x)=3x^2-3sqrt2x+1
Answer :
Thus, the required quadratic polynomial is 3x^2-3sqrt2x+1.
iii) Let the required quadratic polynomial is p(x)=ax^2+bx+c and its zeroes are alpha and beta. Then we have (according to question)
alpha+beta=-b/a=-0/1=0
alpha times beta=c/a=sqrt5=sqrt5/1
a=1, b=0, and c=sqrt5
p(x)=x^2+0.x+sqrt5
p(x)=x^2+sqrt5
Answer :
Thus, the required quadratic polynomial is x^2+sqrt5.
iv) Let the required quadratic polynomial is p(x)=ax^2+bx+c and its zeroes are alpha and beta. Then we have (according to question)
alpha+beta=-b/a=1=-(-1)/1.
alpha times beta= c/a =1=1/1
a=1, b=1, and c=1
p(x)=x^2-x+1
Answer :
Thus, the required quadratic polynomial is x^2-x+1.
v) Let the required quadratic polynomial is p(x)=ax^2+bx+c and its zeroes are alpha and beta. Then we have (according to question)
alpha+beta=-b/a=-1/4
alpha times beta=c/a=1/4
a=4, b=1, and c=1
p(x)=4x^2+x+1
Answer :
Thus, the required quadratic polynomial is 4x^2+x+1
vi) Let the required quadratic polynomial is p(x)=ax^2+bx+c and its zeroes are alpha and beta. Then we have (according to question)
alpha+beta=-b/a=4=-((-4)/1)
alpha times beta=c/a=1=1/1
a=1, b=-4, and c=1
p(x)=x^2-4x+1
Answer :
Thus, the required quadratic polynomial is x^2-4x+1.
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