AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.3 Textbook Questions and Answers.
AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.3
10th Class Maths 3rd Lesson Polynomials Ex 3.3 Textbook Questions and Answers
Question 1.
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
Answer:
i) Given polynomial is x2 – 2x – 8
We have x2 – 2x – 8 = x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
So, the value of x2 – 2x – 8 is zero
when x – 4 = 0 or x + 2 = 0 i.e.,
when x = 4 or x = -2
So, the zeroes of x2 – 2x – 8 are 4 and -2.
Sum of the zeroes = 4 – 2 = 2 Coefficient of ,x -(-2)
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-2)}{1}\) = 2
And product of the zeroes = 4 × (-2) = -8
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-8}{1}\) = -8
ii) Given polynomial is 4s2 – 4s + 1
We have, 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
= 2s (2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
= (2s – 1)2
So, the value of 4s2 – 4s + 1 is zero
when 2s-1 = 0 or s = \(\frac{1}{2}\)
∴ Zeroes of the polynomial are \(\frac{1}{2}\) and \(\frac{1}{2}\)
∴ Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1.
= – \(\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}\) = –\(\frac{-4}{4}\) = 1
And product of the zeroes = \(\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)\) = \(\frac{1}{4}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{1}{4}\)
iii) Given polynomial is 6x2 – 3 – 7x
We have, 6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
The value of 6x2 – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0
i.e., when 3x + 1 = 0 and 2x – 3 = 0
3x = -1 and 2x = 3
x = \(\frac{-1}{3}\) and x = \(\frac{3}{2}\)
∴ The zeroes of 6x2 – 3 – 7x = \(\frac{-1}{3}\) and \(\frac{3}{2}\)
∴ Sum of the zeroes = \(\frac{1}{3}\) + \(\frac{3}{2}\) = \(\frac{7}{6}\).
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-7)}{6}\) = \(\frac{7}{6}\)
And product of the zeroes = \(\left(\frac{-1}{3}\right) \times\left(\frac{3}{2}\right)\) = \(\frac{-1}{2}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\)
iv) Given polynomial is 4u2 + 8u
We have, 4u2 + 8u = 4u (u + 2)
The value of 4u2 + 8u is 0,
when the value of 4u(u + 2) = 0, i.e.,
when u = 0 or u + 2 = 0, i.e.,
when u = 0 (or) u = – 2
∴ The zeroes of 4u2 + 8u are 0 and – 2.
Therefore, sum of the zeroes = 0 + (-2) = -2
= – \(\frac{\text { Coefficient of } u}{\text { Coefficient of } u^{2}}\) = \(\frac{-8}{4}\) = -2
And product of the zeroes 0 . (-2) = 0
= \(\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\) = \(\frac{0}{4}\) = 0
v) Given polynomial is t2 – 15.
We have, t2 – 15 = (t – √15 ) (t + √l5)
The value of t2 – 15 is 0,
when the value of (t – √15 ) (t + √l5) = 0, i.e.,
when t – √15 = 0 or t + √15 = 0, i.e.,
when t = √15 (or) t = -√15
∴ The zeroes of t2 – 15 are √15 and -√15.
Therefore, sum of the zeroes = √15 + (-√15) = 0
= – \(\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{2}}\) = –\(\frac{0}{1}\) = 0
And product of the zeroes √15 × (-√15) = -15
= \(\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\) = \(\frac{-15}{1}\) = -15
vi) Given polynomial is 3x2 – x – 4
we have, 3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)
The value of 3x2 – x – 4 is 0 when the value of (x + 1) (3x – 4) is 0.
i.e., when x + 1 = 0 or 3x – 4 = 0
i.e., when x = -1 or x = \(\frac{4}{3}\)
∴ The zeroes of 3x2 – x – 4 are -1 and \(\frac{4}{3}\)
Therefore, sum of the zeroes = -1 + \(\frac{4}{3}\) = \(\frac{-3+4}{3}\) = \(\frac{1}{3}\)
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-1)}{3}\) = \(\frac{1}{3}\)
And product of the zeroes -1 × \(\frac{4}{3}\) = \(\frac{-4}{3}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-4}{3}\)
Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.
i) \(\frac{1}{4}\), -1
ii) √2, \(\frac{1}{3}\)
iii) 0, √5
iv) 1, 1
v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
vi) 4, 1
Answer:
Let the polynomial be ax2 + bx + c
and its zeroes be α and β.
i) Here, α + β = \(\frac{1}{4}\) and αβ = -1
Thus, the polynomial formed = x2 – (sum of the zeroes)x + product of the zeroes
= x2 – (\(\frac{1}{4}\))x – 1
= x2 – \(\frac{x}{4}\) – 1
The other polynomials are (x2 – \(\frac{x}{4}\) – 1)
then the polynomial is 4x2 – x – 4.
ii) Here, α + β = √2 and αβ = \(\frac{1}{3}\)
Thus, the polynomial formed = x2 – (sum of the zeroes)x + product of the zeroes
= x2 – (√2)x + \(\frac{1}{3}\)
= x2 – √2x + \(\frac{1}{3}\)
The other polynomials are (x2 – √2x + \(\frac{1}{3}\))
then the polynomial is 3x2 – 3√2x + 1.
iii) Here, α + β = 0 and αβ = √5
Thus, the polynomial formed = x2 – (sum of the zeroes)x + product of the zeroes
= x2 – (0)x + √5
= x2 + √5
iv) Let the polynomial be ax2 + bx + c and its zeroes be α and β.
Then α + β = 1 = \(\frac{-(-1)}{1}\) = \(\frac{-b}{a}\) and
αβ = 1 = latex]\frac{1}{1}[/latex] = \(\frac{c}{a}\)
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x2 + x + 1.
v) Let the polynomial be ax2 + bx + c and its zeroes be α and β.
Then α + β = \(\frac{-1}{4}\) = \(\frac{-b}{a}\) and
αβ = \(\frac{1}{4}\) = \(\frac{c}{a}\)
If a = 4, then b = 1 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is 4x2 + x + 1.
vi) Let the polynomial be ax2 + bx + c and its zeroes be α and β.
Then α + β = 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\) and
αβ = 1 = \(\frac{1}{1}\) = \(\frac{c}{a}\)
If a = 1, then b = -4 and c = 1
∴ One quadratic polynomial which satisfies the given conditions is x2 – 4x + 1.
Question 3.
Find the quadratic polynomial, for the zeroes α, β given in each case.
i) 2, -1
ii) √3, -√3
iii) \(\frac{1}{4}\), -1
iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Answer:
i) Let the polynomial be ax2 + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = 2 and β = – 1
Sum of the zeroes = α + β = 2 + (-l) = 1
Product of the zeroes = αβ = 2 × (-1) = -2
Therefore the quadratic polynomial ax2 + bx + c is x2 – (α + β)x + αβ = [x2 – x – 2]
the quadratic polynomial will be x2 – x – 2.
ii) Let the zeroes be α = √3 and β = -√3
Sum of the zeroes = α + β
= √3 + (-√3) = 0
Product of the zeroes = αβ
= √3 × (-√3) = -3
∴ The quadratic polynomial
ax2 + bx + c is [x2 – (α + β)x + αβ]
= [x2 – 0.x + (-3)] = [x2 – 3]
the quadratic polynomial will be x2 – 3.
iii) Let the zeroes be α = \(\frac{1}{4}\) and β = -1
Sum of the zeroes = α + β
= \(\frac{1}{4}\) + (-1) = \(\frac{1+(-4)}{4}\) = \(\frac{-3}{4}\)
Product of the zeroes = αβ
= \(\frac{1}{4}\) × (-1) = \(\frac{-1}{4}\)
∴ The quadratic polynomial
ax2 + bx + c is [x2 – (α + β)x + αβ]
= [x2 – \(\left(\frac{-3}{4}\right)\).x + (\(\frac{-1}{4}\))]
the quadratic polynomial will be 4x2 + 3x – 1.
iv) Let the zeroes be α = \(\frac{1}{2}\) and β = \(\frac{3}{2}\)
Sum of the zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{1+3}{2}\) = \(\frac{4}{2}\) = 2
Product of the zeroes = αβ
= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)
∴ The quadratic polynomial
ax2 + bx + c is [x2 – (α + β)x + αβ]
= [x2 – 2x + (\(\frac{3}{4}\))]
the quadratic polynomial will be 4x2 – 8x + 3.
Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x3 + 3x2 – x – 3 and check the relationship between zeroes and the coefficients.
Answer:
Given cubic polynomial
p(x) = x3 + 3x2 – x – 3
Comparing the given polynomial with ax3 + bx2 + cx + d, we get a = 1, b = 3, c = -1, d = -3
Futher given zeroes are 1,-1 and – 3
p(1) = (1)3 + 3(1)2 – 1 – 3
= 1 + 3 – 1 – 3 = 0
p(-1) = (-1)3 + 3(-1)2 – 1 – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)3 + 3(-3)2 – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore, 1, -1 and -3 are the zeroes of x3 + 3x2 – x – 3.
So, we take α = 1, β = -1 and γ = -3 Now,
α + β + γ = 1 + (-1) + (-3) = -3
αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1
= -1 + 3 – 3 = -1
= \(\frac{c}{a}\) = \(\frac{-1}{1}\) = -1
αβγ = 1 (-1) (-3) = 3 = \(\frac{-d}{a}\) = \(\frac{-(-3)}{1}\) = 3
