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VOOZH | about |
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VOOZH | about |
The Polynomials are fundamental algebraic expressions consisting of variables raised to the whole-number powers and combined using addition, subtraction, and multiplication. They form the basis of many concepts in algebra including solving equations, graphing functions, and mathematical modeling. In this chapter, we will explore polynomials in detail and solve various problems to understand their properties and applications better. Exercise 2.1 | Set 1 focuses on solving the fundamental problems to build a strong foundation in polynomials.
A polynomial is an algebraic expression that involves a sum of powers of one or more variables multiplied by the coefficients. For example: 3x2+2x−5 is a polynomial in the variable x. The degree of a polynomial is the highest power of the variable in the expression. Polynomials can be classified into different types based on their degree and number of terms such as monomials, binomials, and trinomials. They are crucial for solving algebraic equations and analyzing mathematical relationships.
Solution:
Given that,
f(x) = x2 – 2x – 8
To find the zeros of the equation, put f(x) = 0
= x2 – 2x – 8 = 0
= x2 – 4x + 2x – 8 = 0
= x(x – 4) + 2(x – 4) = 0
= (x – 4)(x + 2) = 0
x = 4 and x = -2
Hence, the zeros of the quadratic equation are 4 and -2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x^2
4 + (-2)= – (-2) / 1
2 = 2
Product of roots = constant / coefficient of x^2
4 x (-2) = (-8) / 1
-8 = -8
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
g(s) = 4s2 – 4s + 1
To find the zeros of the equation, put g(s) = 0
= 4s2 – 4s + 1 = 0
= 4s2 – 2s – 2s + 1= 0
= 2s(2s – 1) – (2s – 1) = 0
= (2s – 1)(2s – 1) = 0
s = 1/2 and s = 1/2
Hence, the zeros of the quadratic equation are 1/2 and 1/2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of s / coefficient of s2
1/2 + 1/2 = – (-4) / 4
1 = 1
Product of roots = constant / coefficient of s2
1/2 x 1/2 = 1/4
1/4 = 1/4
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
h(t) = t2 – 15 = t2 +(0)t – 15
To find the zeros of the equation, put h(t) = 0
= t2 – 15 = 0
= (t + √15)(t – √15)= 0
t = √15 and t = -√15
Hence, the zeros of the quadratic equation are √15 and -√15.
Now, Verification
As we know that,
Sum of zeros = – coefficient of t / coefficient of t2
√15 + (-√15) = – (0) / 1
0 = 0
Product of roots = constant / coefficient of t2
√15 x (-√15) = -15/1
-15 = -15
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
f(x) = 6x2 – 3 – 7x
To find the zeros of the equation, we put f(x) = 0
= 6x2 – 3 – 7x = 0
= 6x2 – 9x + 2x – 3 = 0
= 3x(2x – 3) + 1(2x – 3) = 0
= (2x – 3)(3x + 1) = 0
x = 3/2 and x = -1/3
Hence, the zeros of the quadratic equation are 3/2 and -1/3.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
3/2 + (-1/3) = – (-7) / 6
7/6 = 7/6
Product of roots = constant / coefficient of x2
3/2 x (-1/3) = (-3) / 6
-1/2 = -1/2
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
p(x) = x2 + 2√2x – 6
To find the zeros of the equation, put p(x) = 0
= x2 + 2√2x – 6 = 0
= x2 + 3√2x – √2x – 6 = 0
= x(x + 3√2) – √2 (x + 3√2) = 0
= (x – √2)(x + 3√2) = 0
x = √2 and x = -3√2
Hence, the zeros of the quadratic equation are √2 and -3√2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
√2 + (-3√2) = – (2√2) / 1
-2√2 = -2√2
Product of roots = constant / coefficient of x2
√2 x (-3√2) = (-6) / 2√2
-3 x 2 = -6/1
-6 = -6
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
q(x) = √3x2 + 10x + 7√3
To find the zeros of the equation, put q(x) = 0
= √3x2 + 10x + 7√3 = 0
= √3x2 + 3x +7x + 7√3x = 0
= √3x(x + √3) + 7 (x + √3) = 0
= (x + √3)(√3x + 7) = 0
x = -√3 and x = -7/√3
Hence, the zeros of the quadratic equation are -√3 and -7/√3.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
-√3 + (-7/√3) = – (10) /√3
(-3-7)/ √3 = -10/√3
-10/ √3 = -10/√3
Product of roots = constant / coefficient of x2
(-√3) x (-7/√3) = (7√3) / √3
7 = 7
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
f(x) = x2 – (√3 + 1)x + √3
To find the zeros of the equation, put f(x) = 0
= x2 – (√3 + 1)x + √3 = 0
= x2 – √3x – x + √3 = 0
= x(x – √3) – 1 (x – √3) = 0
= (x – √3)(x – 1) = 0
x = √3 and x = 1
Hence, the zeros of the quadratic equation are √3 and 1.
Now, Verification
Sum of zeros = – coefficient of x / coefficient of x2
√3 + 1 = – (-(√3 +1)) / 1
√3 + 1 = √3 +1
Product of roots = constant / coefficient of x2
1 x √3 = √3 / 1
√3 = √3
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
g(x) = a(x2+1)–x(a2+1)
To find the zeros of the equation put g(x) = 0
= a(x2+1)–x(a2+1) = 0
= ax2 + a − a2x – x = 0
= ax2 − a2x – x + a = 0
= ax(x − a) − 1(x – a) = 0
= (x – a)(ax – 1) = 0
x = a and x = 1/a
Hence, the zeros of the quadratic equation are a and 1/a.
Now, Verification :
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
a + 1/a = – (-(a2 + 1)) / a
(a^2 + 1)/a = (a2 + 1)/a
Product of roots = constant / coefficient of x2
a x 1/a = a / a
1 = 1
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
h(s) = 2s2 – (1 + 2√2)s + √2
To find the zeros of the equation put h(s) = 0
= 2s2 – (1 + 2√2)s + √2 = 0
= 2s2 – 2√2s – s + √2 = 0
= 2s(s – √2) -1(s – √2) = 0
= (2s – 1)(s – √2) = 0
x = √2 and x = 1/2
Hence, the zeros of the quadratic equation are √3 and 1.
Now, Verification
As we know that,
Sum of zeros = – coefficient of s / coefficient of s2
√2 + 1/2 = – (-(1 + 2√2)) / 2
(2√2 + 1)/2 = (2√2 +1)/2
Product of roots = constant / coefficient of s2
1/2 x √2 = √2 / 2
√2 / 2 = √2 / 2
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
f(v) = v2 + 4√3v – 15
To find the zeros of the equation put f(v) = 0
= v2 + 4√3v – 15 = 0
= v2 + 5√3v – √3v – 15 = 0
= v(v + 5√3) – √3 (v + 5√3) = 0
= (v – √3)(v + 5√3) = 0
v = √3 and v = -5√3
Hence, the zeros of the quadratic equation are √3 and -5√3.
Now, for verification
Sum of zeros = – coefficient of v / coefficient of v2
√3 + (-5√3) = – (4√3) / 1
-4√3 = -4√3
Product of roots = constant / coefficient of v2
√3 x (-5√3) = (-15) / 1
-5 x 3 = -15
-15 = -15
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
p(y) = y2 + (3√5/2)y – 5
To find the zeros of the equation put f(v) = 0
= y2 + (3√5/2)y – 5 = 0
= y2 – √5/2 y + 2√5y – 5 = 0
= y(y – √5/2) + 2√5 (y – √5/2) = 0
= (y + 2√5)(y – √5/2) = 0
This gives us 2 zeros,
y = √5/2 and y = -2√5
Hence, the zeros of the quadratic equation are √5/2 and -2√5.
Now, Verification
As we know that,
Sum of zeros = – coefficient of y / coefficient of y2
√5/2 + (-2√5) = – (3√5/2) / 1
-3√5/2 = -3√5/2
Product of roots = constant / coefficient of y2
√5/2 x (-2√5) = (-5) / 1
– (√5)2 = -5
-5 = -5
Hence the relationship between zeros and their coefficients are verified.
Solution:
Given that,
q(y) = 7y2 – (11/3)y – 2/3
To find the zeros of the equation put q(y) = 0
= 7y2 – (11/3)y – 2/3 = 0
= (21y2 – 11y -2)/3 = 0
= 21y2 – 11y – 2 = 0
= 21y2 – 14y + 3y – 2 = 0
= 7y(3y – 2) – 1(3y + 2) = 0
= (3y – 2)(7y + 1) = 0
y = 2/3 and y = -1/7
Hence, the zeros of the quadratic equation are 2/3 and -1/7.
Now, Verification
As we know that,
Sum of zeros = – coefficient of y / coefficient of y2
2/3 + (-1/7) = – (-11/3) / 7
-11/21 = -11/21
Product of roots = constant / coefficient of y2
2/3 x (-1/7) = (-2/3) / 7
– 2/21 = -2/21
Hence the relationship between zeros and their coefficients are verified.
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -8/3 and
Product of zero = 4/3
Therefore,
Required polynomial f(x) is,
= x2 – (-8/3)x + (4/3)
= x2 + 8/3x + (4/3)
To find the zeros we put f(x) = 0
= x2 + 8/3x + (4/3) = 0
= 3x2 + 8x + 4 = 0
= 3x2 + 6x + 2x + 4 = 0
= 3x(x + 2) + 2(x + 2) = 0
= (x + 2) (3x + 2) = 0
= (x + 2) = 0 and, or (3x + 2) = 0
Hence, the two zeros are -2 and -2/3.
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = 21/8 and
Product of zero = 5/16
Therefore,
The required polynomial f(x) is,
= x2 – (21/8)x + (5/16)
= x2 – 21/8x + 5/16
To find the zeros we put f(x) = 0
= x2 – 21/8x + 5/16 = 0
= 16x2 – 42x + 5 = 0
= 16x2 – 40x – 2x + 5 = 0
= 8x(2x – 5) – 1(2x – 5) = 0
= (2x – 5) (8x – 1) = 0
= (2x – 5) = 0 and, or (8x – 1) = 0
Hence, the two zeros are 5/2 and 1/8.
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -2√3 and
Product of zero = -9
Therefore,
The required polynomial f(x) is,
= x2 – (-2√3)x + (-9)
= x2 + 2√3x – 9
To find the zeros we put f(x) = 0
= x2 + 2√3x – 9 = 0
= x2 + 3√3x – √3x – 9 = 0
= x(x + 3√3) – √3(x + 3√3) = 0
= (x + 3√3) (x – √3) = 0
= (x + 3√3) = 0 and, or (x – √3) = 0
Hence, the two zeros are -3√3and √3.
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -3/2√5 and
Product of zero = -1/2
Therefore,
The required polynomial f(x) is,
= x2 – (-3/2√5)x + (-1/2)
= x2 + 3/2√5x – 1/2
To find the zeros we put f(x) = 0
= x2 + 3/2√5x – 1/2 = 0
= 2√5x2 + 3x – √5 = 0
= 2√5x2 + 5x – 2x – √5 = 0
= √5x(2x + √5) – 1(2x + √5) = 0
= (2x + √5) (√5x – 1) = 0
= (2x + √5) = 0 and, or (√5x – 1) = 0
Hence, the two zeros are -√5/2 and 1/√5.
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(x) where a = 1, b = -5 and c = 4
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-5)/1 = 5,
Product of the roots = αβ = c/a = 4/1 = 4
We have to find 1/α +1/β – 2αβ
= [(α +β)/ αβ] – 2αβ
= (5)/ 4 – 2(4) = 5/4 – 8 = -27/ 4
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1,
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-7)/5 = 7/5
Product of the roots = αβ = c/a = 1/5
We have to find 1/α +1/β
= (α +β)/ αβ
= (7/5)/ (1/5) = 7
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = – 4
So, we can find,
Sum of the roots = α+β = -b/a = – (-1)/1 = 1
Product of the roots = αβ = c/a = -4 /1 = – 4
We have to find, 1/α +1/β – αβ
= [(α +β)/ αβ] – αβ
= [(1)/ (-4)] – (-4) = -1/4 + 4 = 15/ 4
Solution:
Given that:
α and β are the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = – 2
So, we can find
Sum of the roots = α+β = -b/a = – (1)/1 = -1,
Product of the roots = αβ = c/a = -2 /1 = – 2
We have to find, 1/α – 1/β
= [(β – α)/ αβ] = [β-α]/(αβ) x (α-β)/αβ = (√(α+β)2 -4αβ) / αβ = √(1+8) / 2 = 3/2
Solution:
Given that,
The quadratic polynomial f(x) where a = 4, b = -8k and c = – 9
And, for roots to be negative of each other, let us assume that the roots α and – α.
Using these values we can find,
Sum of the roots = α – α = -b/a = – (-8k)/1 = 8k = 0 [∵ α – α = 0]
= k = 0
Solution:
Given that,
The quadratic polynomial f(t)=kt2 + 2t + 3k, where a = k, b = 2 and c = 3k ,
Sum of the roots = Product of the roots
= (-b/a) = (c/a)
= (-2/k) = (3k/k)
= (-2/k) = 3
Hence k = -2/3
Solution:
Given that,
α and β are the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-5)/4 = 5/4
Product of the roots = αβ = c/a = -1/4
We have to find, α^2 β+α β^2
= αβ(α +β)
= (-1/4)(5/4) = -5/16
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(t) where a = 1, b = -4 and c = 3
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-4)/1 = 4 ,
Product of the roots = αβ = c/a = 3/1 = 3
We have to find, α4 β3 + α3 β4
= α3 β3 (α +β)
= (αβ)3 (α +β)
= (3)3 (4) = 27 x 4 = 108
Read More:
Exercise 2.1 | Set 1 of RD Sharma's Class 10 Mathematics textbook introduces the basic concepts of polynomials. This section covers the definition of polynomials, their classification based on degree and number of terms, identification of polynomials, and basic operations with polynomials. Students learn to recognize polynomials in different forms, determine their degrees, identify coefficients and constants, and understand the concept of zeros of polynomials. The exercise also includes problems on evaluating polynomials for given values and finding the values of variables for which a polynomial equals zero.
