Class10 NCRT Polynomials Exercise – 2.2 pdf || UP Board
Polynomials
Exercise – 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2-2x-8
Solution:
By the
factorization method,
x2-4x+2x-8 = 0
x(x-4) +2(x-4) = 0
(x+2) (x-4) = 0
x = -2, 4
So, zeroes α=-2, β=4
Verification,
-2+4 = -(-2)/1
2 = 2
-2×4 = -8/1
-8 = -8
The
relation between the coefficients is true in both the cases.
(ii) 4s2-4s+1
Solution:
By the
factorization method,
4s2-2s-2s+1 = 0
2s(2s-1) -1(2s-1) = 0
(2s-1) (2s-1) = 0
s = 1/2, 1/2
So, zeroes α= 1/2, β= 1/2
Verification,
sum of zeroes(α+β) =(-b )/a
½ + 1/2 = -(-4/4)
2/2 = 4/4
1 = 1
product of zeroes (αβ) =(c )/a
½ × 1/2 = 1/4
1/4 = 1/4
The
relation between the coefficients is true in both the cases.
(iii) 6x2-3-7x
Solution:
By the
factorization method,
6x2-9x+2x-3 = 0
3x(2x-3) +1(2x-3) = 0
(3x+1) (2x-3) = 0
x = -1/3, 3/2
So, zeroes α=
3/2, β=
-1/3
Verification,
sum of zeroes(α+β) =(-b )/a
3/2 + -1/3 = -(-7)/6
(9-2)/6 = 7/6
7/6 = 7/6
product of zeroes (αβ) =(c )/a
3/2 × -1/3 = -3/6
-3/6 = -3/6
The
relation between the coefficients is true in both the cases.
(iv) 4u2+8u
Solution:
By the
factorization method,
4u(u+2) = 0
u = 0, -2
So, zeroes α= 0, β= -2
Verification,
sum of zeroes (α+β) =(-b )/a
0 + -2 = -8/4
-2 = -2
product of zeroes (αβ) =(c )/a
0 × -2 = 0/4
0 = 0
The
relation between the coefficients is true in both the cases.
(v) t2-15
Solution:
By the
factorization method,
t2 - 15 = 0
t = ±√15
t = √15, -√15
So, zeroes α= √15, β= -√15
Verification,
sum of zeroes (α+β) =(-b )/a
√15 + -√15 = -(0)/1
0 = 0
product of zeroes (αβ) =(c )/a
√15 × -√15 = -15
-15 = -15
The
relation between the coefficients is true in both the cases.
(vi) 3x2-x-4
Solution:
By the
factorization method,
3x2-4x+3x-4 = 0
x(3x-4) +1(3x-4) = 0
(3x-4) (x+1) = 0
x = 4/3, -1
So, zeroes α= 4/3, β= -1
Verification,
sum of zeroes (α+β) =(-b )/a
4/3 + -1 = -(-1)/3
(4-3)/3 = 1/3
1/3 = 1/3
product of zeroes (αβ) =(c )/a
4/3 × -1 = -4/3
-4/3 = -4/3
The
relation between the coefficients is true in both the cases.
2. Find
a quadratic polynomial each with the given numbers as the sum and product of
its zeroes respectively:
(i) ¼,
-1
Solution:
Let a
quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.
According
to Question,
sum of zeroes α + β = -b/a =
¼
-b/a = ¼
by
comparison,
-b = 1, a = 4
b = -1, a = 4
product
of zeros
αβ = c/a =
-1
c/a = -1
or c/4 = -1 (a
= 4)
c =
-4
Now,
b = -1,
a = 4,
c = -4
Hence,
ax2+bx+c
4x2-1x-4
The
quadratic polynomial is 4x2-1x-4.
(ii) √2,
1/3
Solution:
Let a
quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.
According
to Question,
sum of zeroes α + β = -b/a =
√2
-b/a = √2
or -b/a
= √2/1
by
comparison,
-b = √2, a = 1
b = -√2, a =
1
product
of zeros αβ = c/a =
1/3
c/a = 1/3
or c/1 = 1/3 (a
= 1)
c =
1/3
Now,
b = -√2, a = 1, c = 1/3
Hence,
ax2+bx+c = 0
3x2-3√2+1
= 0
The quadratic polynomial is 3x2-3√2+1.
हल:
Let a
quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.
According
to Question,
sum of zeroes α + β = -b/a =
0
-b/a = 0
or -b/a
= 0/1
by
comparison,
b = 0, a = 1
product
of zeros αβ = c/a =
√5
c/a = √5/1
or c/1 = √5/1 (a
= 1)
c =
√5
Now,
b = 0, a = 1, c = √5
Hence,
ax2+bx+c
= 0
x2-0x+√5 = 0
x2+√5
= 0
The
quadratic polynomial is x2+√5.
(iv) 1,
1
हल:
Let a
quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.
According
to Question,
sum of zeroes α + β = -b/a =
1
-b/a = 1
or -b/a
= 1/1
by
comparison,
b = -1, a = 1
product
of zeros αβ = c/a =
1
c/a = 1/1
or c/1 = 1/1 (a
= 1)
c =
1
Now,
b = -1, a = 1, c = 1
Hence,
ax2+bx+c
= 0
x2-1x+1 = 0
The
quadratic polynomial is x2-1x+1.
(v) -1/4,
1/4
हल:
Let a
quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.
According
to Question,
sum of zeroes α + β = -b/a =
-1/4
-b/a = -1/4
or b/a
= 1/4
by
comparison,
b = 1, a = 4
product
of zeros αβ = c/a =
1/4
c/a = 1/4
or c/4 = 1/4 (a
= 4)
c =
1
Now,
b = 1, a = 4, c = 1
Hence,
ax2+bx+c
= 0
4x2+1x+1 = 0
The
quadratic polynomial is 4x2+1x+1.
(vi) 4,
1
हल:
Let a
quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.
According
to Question,
sum of zeroes α + β = -b/a =
4
-b/a = 4
or b/a
= -4/1
by
comparison,
b = -4, a = 1
product
of zeros αβ = c/a =
1
c/a = 1
or c/1 = 1/1 (a
= 1)
c
= 1
Now,
b = -4, a = 1, c = 1
Hence,
ax2+bx+c
= 0
x2-4x+1 = 0
The
quadratic polynomial is x2-4x+1.
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