RS Aggarwal Solutions Chapter 2 Polynomials Exercise - 2A Class 10 Maths
Chapter Name | RS Aggarwal Chapter 2 Polynomials |
Book Name | RS Aggarwal Mathematics for Class 10 |
Other Exercises |
|
Related Study | NCERT Solutions for Class 10 Maths |
Exercise 2A Solutions
1. Find the zeros of the polynomial f(x) = x2 + 7x + 12 and verify the relation between its zeroes and coefficients.
Solution
x2 + 7x + 12 = 0
⇒ x2 + 4x + 3x + 12 = 0
⇒ x(x + 4) + 3(x + 4) = 0
⇒ (x + 4) (x + 3) = 0
⇒ (x + 4) = 0 or (x + 3) = 0
⇒ x = −4 or x = −3
Sum of zeroes = −4 + (−3) = -7/1 = -coefficient of x/coefficient of x2
Product of zeroes = (−4) (−3) = 12/1 = constant term/coefficient of x2
2. Find the zeroes of the polynomial f(x) = x2 ˗ 2x ˗ 8 and verify the relation between its zeroes and coefficients.
Solution
x2˗ 2x ˗ 8 = 0
⇒ x2˗ 4x + 2x ˗ 8 = 0
⇒ x(x ˗ 4) + 2(x ˗ 4) = 0
⇒ (x ˗ 4) (x + 2) = 0
⇒ (x ˗ 4) = 0 or (x + 2) = 0
⇒ x = 4 or x = −2
Sum of zeroes = 4 + (−2) = 2 = 2/1 = -(coefficient of x)/(coefficient of x2)
Product of zeros = (4)(-2) = -8/1 = (constant term)/(coefficient of x2)
3. Find the zeroes of the quadratic polynomial f(x) = x2 + 3x ˗ 10 and verify the relation between its zeroes and coefficients.
Solution
We have:
f(x) = x2 + 3x ˗ 10
= x2 + 5x ˗ 2x ˗ 10
= x(x + 5) ˗ 2(x + 5)
= (x ˗ 2) (x + 5)
∴ f(x) = 0
⇒ (x ˗ 2) (x + 5) = 0
⇒ x ˗ 2 = 0 or x + 5 = 0
⇒ x = 2 or x = −5.
So, the zeroes of f(x) are 2 and −5.
Sum of zeroes = 2 + (−5) = −3 = -3/1 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = 2 × (−5) = −10 = -10/1 = (constant term)/(coefficient of x2)
4. Find the zeroes of the quadratic polynomial f(x) = 4x2˗ 4x ˗ 3 and verify the relation between its zeroes and coefficients.
Solution
We have:
f(x) = 4x2˗ 4x ˗ 3
= 4x2˗ (6x ˗ 2x) ˗ 3
= 4x2˗ 6x + 2x ˗ 3
= 2x (2x ˗ 3) + 1(2x ˗ 3)
= (2x + 1) (2x ˗ 3)
∴ f(x) = 0
⇒ (2x + 1) (2x ˗ 3) = 0
⇒ 2x + 1 = 0 or 2x ˗ 3 = 0
⇒ x = -1/2 or x = 3/2
So, the zeroes of f(x) are -1/2 and 3/2.
Sum of zeroes = (-1/2) + (3/2)
= (-1 + 3)/2
= 2/2
= 1 = -(coefficient of x)/(coefficient of x2)
Product of zeros = (-1/2) × (3/2)
= -3/4 = (constant term)/(coefficient of x2)
5. Find the zeroes of the quadratic polynomial f(x) = 5x2˗ 4 ˗ 8x and verify the relationship between the zeroes and coefficients of the given polynomial.
Solution
We have,
f(x) = 5x2˗ 4 ˗ 8x
= 5x2 ˗ 8x ˗ 4
= 5x2˗ (10x ˗ 2x) ˗ 4
= 5x2˗ 10x + 2x ˗ 4
= 5x (x ˗ 2) + 2(x ˗ 2)
= (5x + 2)(x ˗ 2)
∴ f(x) = 0
⇒ (5x + 2) (x ˗ 2) = 0
⇒ 5x + 2 = 0 or x ˗ 2 = 0
⇒ x = -2/5 or x = 2
So, the zeroes of f(x) are -2/5 and 2.
Sum of zeroes = (-2/5) + 2
= (-2 + 10)/5
= 8/5 = -(coefficient of x)/(coefficient of x2)
Product of zeros = (-2/5 × 2)
= -4/5 = (constant term)/(coefficient of x2)
6. Find the zeroes of the polynomial f(x) = 2√3x2 -5x + √3 and verify the relation between its zeroes and coefficients.
Solution
2√3x2 ˗ 5x + √3
⇒ 2√3x2 ˗ 2x ˗ 3x + √3
⇒ 2x (√3x ˗ 1) ˗ √3 (√3x ˗ 1) = 0
⇒ (√3x ˗ 1) or (2x− √3) = 0
⇒ (√3x ˗ 1) = 0 or (2x − √3) = 0
⇒ x = 1/√3 or x = √3/2
⇒ x = 1/√3 × 1/√3 = √3/3 or x = √3/2
Sum of zeros = √3/3 + √3/2 + (5√3)/6 = -(coefficient of x)/(coefficient of x2)
Product of zeros = √3/3 × √3/2 = √3/6 = (constant term)/(coefficient of x2)
7. Find the zeroes of the quadratic polynomial 2x2 - 11x + 15 and verify the relation between the zeroes and the coefficients.
Solution
f(x) = 2x2˗ 11x + 15
= 2x2˗ (6x + 5x) + 15
= 2x2˗ 6x ˗ 5x + 15
= 2x(x ˗ 3) ˗ 5(x ˗ 3)
= (2x ˗ 5)(x ˗ 3)
∴ f(x) = 0
⇒ (2x ˗ 5) (x ˗ 3) = 0
⇒ 2x ˗ 5 = 0 or x ˗ 3 = 0
⇒ x = 5/2 or x = 3
So, the zeroes of f(x) are 5/2 and 3.
Sum of zeros = 5/2 + 3
= (5 + 6)/2
= 11/2 = -(coefficient of x)/(coefficient of x2)
Product of zeros = 5/2 × 3
= -15/2 = (constant term)/(coefficient of x2)
8. Find the zeroes of the quadratic polynomial 4x2˗ 4x + 1 and verify the relation between the zeroes and the coefficients.
Solution
4x2 ˗ 4x + 1 = 0
⇒ (2x)2 ˗ 2(2x)(1) + (1)2 = 0
⇒ (2x ˗ 1)2 = 0 [∵ a2 – 2ab + b2 = (a–b)2]
⇒ (2x ˗ 1)2 = 0
⇒ x = 1/2 or x = 1/2
Sum of zeroes = 1/2 + 1/2
= 1
= 1/1 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = 1/2 × 1/2
= 1/4 = (constant term)/(coefficient of x2)
9. Find the zeroes of the quadratic polynomial (x2 ˗ 5) and verify the relation between the zeroes and the coefficients.
Solution
We have:
f(x) = x2˗ 5
It can be written as x2 + 0x ˗ 5.
= (x2 − (√5)2)
= (x + √5) (x − √5)
∴ f(x) = 0 ⇒ (x + √5) (x − √5) = 0
⇒ x + √5 = 0 or x − √5 = 0
⇒ x = −√5 or x = √5
So, the zeroes of f(x) are −√5 and √5.
Here, the coefficient of x is 0 and the coefficient of x2 is 1.
Sum of zeroes = -√5 + √5
= 1
= 0/1 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = -√5 × √5
= -5/1 = (constant term)/(coefficient of x2)
10. Find the zeroes of the quadratic polynomial (8x2˗ 4) and verify the relation between the zeroes and the coefficients.
Solution
We have:
f(x) = 8x2 ˗ 4
It can be written as 8x2 + 0x ˗ 4
= 4{(√2x)2˗ (1)2}
= 4(√2x + 1)(√2x ˗ 1)
∴ f(x) = 0
⇒ (√2x + 1) (√2x ˗ 1) = 0
⇒ (√2x + 1) = 0 or √2x ˗ 1 = 0
⇒ x = -1/√2 or x = 1/√2
So, the zeroes of f(x) are -1/√2 and 1/√2
Here the coefficient of x is 0 and the coefficient of x2 is √2
Sum of zeroes = -1/√2 + 1/√2
= (-1 + 1)/√2
= 0/√2 = -(coefficient of x)/(coefficient of x2)
Product of zeros = -1/√2 × 1/√2
= (-1 × 4)/(2 × 4)
= -4/8 = (constant term/(coefficient of x2)
11. Find the zeroes of the quadratic polynomial (5y2 + 10y) and verify the relation between the zeroes and the coefficients.
Solution
We have,
f (u) = 5u2 + 10u
It can be written as 5u (u + 2)
∴ f (u) = 0
⇒ 5u = 0 or u + 2 = 0
⇒ u = 0 or u = −2
So, the zeroes of f (u) are −2 and 0.
Sum of the zeroes = −2 + 0
= −2 = (-2 × 5)/(1 × 5)
= -10/5 = (-coefficient of x)/(coefficient of u2)
Product of zeros = -2 × 0
= 0
= (0 × 5)/(1 × 5)
= -0/5 = (constant term)/(-coefficient of u2)
12. Find the zeroes of the quadratic polynomial (3x2˗ x ˗ 4) and verify the relation between the zeroes and the coefficients.
Solution
3x2˗ x ˗ 4 = 0
⇒ 3x2˗ 4x + 3x ˗ 4 = 0
⇒ x (3x ˗ 4) + 1 (3x ˗ 4) = 0
⇒ (3x ˗ 4) (x + 1) = 0
⇒ (3x ˗ 4) or (x + 1) = 0
⇒ x = 4/3 or x = ˗ 1
Sum of zeroes = 4/3 + (-1)
= 1/3 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = 4/3 × (-1)
= -4/3 = (constant term)/(coefficient of x2)
13. Find the quadratic polynomial whose zeroes are 2 and -6. Verify the relation between the coefficients and the zeroes of the polynomial.
Solution
Let α = 2 and β = -6
Sum of the zeroes, (α + β) = 2 + (-6) = -4
Product of the zeros, αβ = 2 × (-6) = -12
∴ Required polynomial = x2 – (α + β)x + αβ
= x2 - (-4x) – 12
= x2 + 4x – 12
Sum of zeros = -4
= -4/1 (-coefficient of x)/(coefficient of x2)
Product of zeros = -12
= -12/1 = (constant term)/(coefficient of x2)
14. Find the quadratic polynomial whose zeroes are 2/3 and -1/4 . Verify the relation between the coefficients and the zeroes of the polynomial.
Solution
Let = α = 2/3 and β = -1/4
Sum of the zeroes = (α + β) = 2/3 + (-1/4)
= (8 – 3)/12
= 5/12
Product of zeroes = αβ = 2/3 × (-1/4)
= -2/12
= -1/6
∴ Required polynomial = x2 – (α + β)x + αβ
= x2 – 5/12x + (-1/6)
= x2 – 5/12x – 1/6
Sum of the zeros = 5/12 = -(coefficient of x)/(coefficient of x2)
Product of zeros = -1/6 = (constant term)/(coefficient of x2)
15. Find the quadratic polynomial, sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.
Solution
Let α and β be the zeroes of the required polynomial f(x).
Then (α + β) = 8 and αβ = 12
∴ f(x) = x2˗ (α + β)x + αβ
⇒ f(x) = x2˗ 8x + 12
Hence, required polynomial f(x) = x2 ˗ 8x + 12
∴ f(x) = 0
⇒ x2 ˗ 8x + 12 = 0
⇒ x2˗ (6x + 2x) + 12 = 0
⇒ x2˗ 6x ˗ 2x + 12 = 0
⇒ x (x – 6) – 2 (x – 6) = 0
⇒ (x – 2) (x – 6) = 0
⇒ (x – 2) = 0 or (x – 6) = 0
⇒ x = 2 or x = 6
So, the zeroes of f(x) are 2 and 6.
16. Find the quadratic polynomial, sum of whose zeroes is 0 and their product is -1. Hence, find the zeroes of the polynomial.
Solution
Let α and β be the zeroes of the required polynomial f(x).
Then (α + β) = 0 and αβ = -1
∴ f(x) = x2 ˗ (α + β)x + αβ
⇒ f(x) = x2˗ 0x + (-1)
⇒ f(x) = x2 ˗ 1
Hence, required polynomial f(x) = x2˗ 1.
∴ f(x) = 0
⇒ x2˗ 1 = 0
⇒ (x + 1) (x – 1) = 0
⇒ (x + 1) = 0 or (x – 1) = 0
⇒ x = -1 or x = 1
So, the zeroes of f(x) are -1 and 1.
17. Find the quadratic polynomial, sum of whose zeroes is (5/2) and their product is 1. Hence, find the zeroes of the polynomial.
Solution
Let α and β be the zeroes of the required polynomial f(x).
Then (α + β) = 5/2 and αβ = 1
∴ f(x) = x2 - (α + β) x + αβ
⇒ f(x) = x2 -5/2x + 1
⇒ f(x) = 2x2 – 5x + 2
Hence, the required polynomial is f(x) = 2x2 – 5x + 2
∴ f(x) = 0 ⇒ 2x2 – 5x + 2 = 0
⇒ 2x2 – (4x + x) + 2 = 0
⇒ 2x2 – 4x – x + 2 = 0
⇒ 2x (x – 2) – 1 (x – 2) = 0
⇒ (2x – 1) (x – 2) = 0
⇒ (2x – 1) = 0 or (x – 2) = 0
⇒ x = 1/2 or x = 2
So, the zeros of f(x) are 1/2 and 2.
18. Find the quadratic polynomial, sum of whose zeroes is √2 and their product is (1/3).
Solution
We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula
x2 – (Sum of the roots)x + Product of roots = 0
⇒ x2 – √2x + 1/3 = 0
⇒ 3x2 – 3√2x + 1 = 0
19. If x = 2/3 and x = -3 are the roots of the quadratic equation ax3 + 2ax + 5x + 10 then find the value of a and b.
Solution
Given: ax2 + 7x + b = 0
Since, x = 2/3 is the root of the above quadratic equation
Hence, it will satisfy the above equation.
Therefore, we will get
a(2/3)2 + 7(2/3) + b = 0
⇒ 4/9a + 14/3 + b = 0
⇒ 4a + 42 + 9b = 0
⇒ 4a + 9b = – 42 …(1)
Since, x = –3 is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
a (–3)2 + 7 (–3) + b = 0
⇒ 9a – 21 + b = 0
⇒ 9a + b = 21 ...(2)
From (1) and (2), we get
a = 3, b = –6
20. If (x + a) is a factor of the polynomial 2x2 + 2ax + 5x + 10, find the value of a.
Solution
Given: (x + a) is a factor of 2x2 + 2ax + 5x + 10
So, we have
x + a = 0
⇒ x = – a
Now, it will satisfy the above polynomial.
Therefore, we will get
2 (–a)2 + 2a(–a) + 5(–a) + 10 = 0
⇒ 2a2 –2a2 – 5a + 10 = 0
⇒ – 5a = – 10
⇒ a = 2
21. One zero of the polynomial 3x3 + 16x2 + 15x – 18 is 2/3. Find the other zeros of the polynomial.
Solution
Given: x = 2/3 is one of the zero of 3x3 + 16x2 + 15x – 18
Now, we have
x = 2/3
⇒ x – 2/3 = 0
Now, we divide 3x3 + 16x2 + 15x – 18 by x – 2/3 to find the quotient
So, the quotient is 3x2 + 18x + 27
Now,
3x2 + 18x + 27 = 0
⇒ 3x2 + 9x + 9x + 27 = 0
⇒ 3x(x + 3) + 9(x + 3) = 0
⇒ (x + 3) (3x + 9) = 0
⇒ (x + 3) = 0 or (3x + 9) = 0
⇒ x = –3 or x = –3