# Class 10 RD Sharma Solutions- Chapter 2 Polynomials – Exercise 2.1 | Set 1

• Last Updated : 02 Feb, 2021

### (i) f(x) = x2 – 2x – 8

Solution:

Given that,

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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.

### (ii) g(s) = 4s2 – 4s + 1

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.

### (iii) h(t)=t2 – 15

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.

### (iv) f(x) = 6x2 – 3 – 7x

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.

### (v) p(x) = x2 + 2√2x – 6

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.

### (vi) q(x)=√3x2 + 10x + 7√3

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.

### (vii) f(x) = x2 – (√3 + 1)x + √3

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.

### (viii) g(x) = a(x2+1)–x(a2+1)

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.

### (ix) h(s) = 2s2 – (1 + 2√2)s + √2

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.

### (x) f(v) = v2 + 4√3v – 15

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.

### (xi) p(y) = y2 + (3√5/2)y – 5

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.

### (xii) q(y) = 7y2 – (11/3)y – 2/3

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.

### (i) -8/3, 4/3

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 = -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.

### (ii) 21/8, 5/16

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.

### (iii) -2√3, -9

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.

### (iv) -3/2√5, -1/2

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.

### Question 3. If α and β are the zeros of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of 1/α + 1/β – 2αβ.

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

### Question 4. If α and β are the zeros of the quadratic polynomial p(y) = 5y2 – 7y + 1, find the value of 1/α+1/β.

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

### Question 5. If α and β are the zeros of the quadratic polynomial f(x)=x2 – x – 4, find the value of 1/α+1/β–αβ.

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

### Question 6. If α and β are the zeroes of the quadratic polynomial f(x) = x2 + x – 2, find the value of 1/α – 1/β.

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

### Question 7. If one of the zero of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is negative of the other, then find the value of k.

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

### Question 8. If the sum of the zeroes of the quadratic polynomial f(t)=kt2 + 2t + 3k is equal to their product, then find the value of k.

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

### Question 9. If α and β are the zeros of the quadratic polynomial p(x) = 4x2 – 5x – 1, find the value of α2 β+α β2

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

### Question 10. If α and β are the zeros of the quadratic polynomial f(t)=t2– 4t + 3, find the value of α4 β3+α3 β4.

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

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