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

• Last Updated : 02 Feb, 2021

### (i) f(x) = x2 â€“ 2x â€“ 8

Solution:

Given that,

f(x) = x2 â€“ 2x â€“ 8

To find the zeros of the equation, put f(x) = 0

= x2 â€“ 2x â€“ 8 = 0

= x2 â€“ 4x + 2x â€“ 8 = 0

= x(x â€“ 4) + 2(x â€“ 4) = 0

= (x â€“ 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 = â€“ coefficient of x / coefficient of x^2

4 + (-2)= â€“ (-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 â€“ 4s + 1

Solution:

Given that,

g(s) = 4s2 â€“ 4s + 1

To find the zeros of the equation, put g(s) = 0

= 4s2 â€“ 4s + 1 = 0

= 4s2 â€“ 2s â€“ 2s + 1= 0

= 2s(2s â€“ 1) â€“ (2s â€“ 1) = 0

= (2s â€“ 1)(2s â€“ 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 = â€“ coefficient of s / coefficient of s2

1/2 + 1/2 = â€“ (-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 â€“ 15

Solution:

Given that,

h(t) = t2 â€“ 15 = t2 +(0)t â€“ 15

To find the zeros of the equation, put h(t) = 0

= t2 â€“ 15 = 0

= (t + âˆš15)(t â€“ âˆš15)= 0

t = âˆš15 and t = -âˆš15

Hence, the zeros of the quadratic equation are âˆš15 and -âˆš15.

Now, Verification

As we know that,

Sum of zeros = â€“ coefficient of t / coefficient of t2

âˆš15 + (-âˆš15) = â€“ (0) / 1

0 = 0

Product of roots = constant / coefficient of t2

âˆš15 x (-âˆš15) = -15/1

-15 = -15

Hence the relationship between zeros and their coefficients are verified.

### (iv) f(x) = 6x2 â€“ 3 â€“ 7x

Solution:

Given that,

f(x) = 6x2 â€“ 3 â€“ 7x

To find the zeros of the equation, we put f(x) = 0

= 6x2 â€“ 3 â€“ 7x = 0

= 6x2 â€“ 9x + 2x â€“ 3 = 0

= 3x(2x â€“ 3) + 1(2x â€“ 3) = 0

= (2x â€“ 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 = â€“ coefficient of x / coefficient of x2

3/2 + (-1/3) = â€“ (-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 + 2âˆš2x â€“ 6

Solution:

Given that,

p(x) = x2 + 2âˆš2x â€“ 6

To find the zeros of the equation, put p(x) = 0

= x2 + 2âˆš2x â€“ 6 = 0

= x2 + 3âˆš2x â€“ âˆš2x â€“ 6 = 0

= x(x + 3âˆš2) â€“ âˆš2 (x + 3âˆš2) = 0

= (x â€“ âˆš2)(x + 3âˆš2) = 0

x = âˆš2 and x = -3âˆš2

Hence, the zeros of the quadratic equation are âˆš2 and -3âˆš2.

Now, Verification

As we know that,

Sum of zeros = â€“ coefficient of x / coefficient of x2

âˆš2 + (-3âˆš2) = â€“ (2âˆš2) / 1

-2âˆš2 = -2âˆš2

Product of roots = constant / coefficient of x2

âˆš2 x (-3âˆš2) = (-6) / 2âˆš2

-3 x 2 = -6/1

-6 = -6

Hence the relationship between zeros and their coefficients are verified.

### (vi) q(x)=âˆš3x2 + 10x + 7âˆš3

Solution:

Given that,

q(x) = âˆš3x2 + 10x + 7âˆš3

To find the zeros of the equation, put q(x) = 0

= âˆš3x2 + 10x + 7âˆš3 = 0

= âˆš3x2 + 3x +7x + 7âˆš3x = 0

= âˆš3x(x + âˆš3) + 7 (x + âˆš3) = 0

= (x + âˆš3)(âˆš3x + 7) = 0

x = -âˆš3 and x = -7/âˆš3

Hence, the zeros of the quadratic equation are -âˆš3 and -7/âˆš3.

Now, Verification

As we know that,

Sum of zeros = â€“ coefficient of x / coefficient of x2

-âˆš3 + (-7/âˆš3) = â€“ (10) /âˆš3

(-3-7)/ âˆš3 = -10/âˆš3

-10/ âˆš3 = -10/âˆš3

Product of roots = constant / coefficient of x2

(-âˆš3) x (-7/âˆš3) = (7âˆš3) / âˆš3

7 = 7

Hence the relationship between zeros and their coefficients are verified.

### (vii) f(x) = x2 â€“ (âˆš3 + 1)x + âˆš3

Solution:

Given that,

f(x) = x2 â€“ (âˆš3 + 1)x + âˆš3

To find the zeros of the equation, put f(x) = 0

= x2 â€“ (âˆš3 + 1)x + âˆš3 = 0

= x2 â€“ âˆš3x â€“ x + âˆš3 = 0

= x(x â€“ âˆš3) â€“ 1 (x â€“ âˆš3) = 0

= (x â€“ âˆš3)(x â€“ 1) = 0

x = âˆš3 and x = 1

Hence, the zeros of the quadratic equation are âˆš3 and 1.

Now, Verification

Sum of zeros = â€“ coefficient of x / coefficient of x2

âˆš3 + 1 = â€“ (-(âˆš3 +1)) / 1

âˆš3 + 1 = âˆš3 +1

Product of roots = constant / coefficient of x2

1 x âˆš3 = âˆš3 / 1

âˆš3 = âˆš3

Hence the relationship between zeros and their coefficients are verified.

### (viii) g(x) = a(x2+1)â€“x(a2+1)

Solution:

Given that,

g(x) = a(x2+1)â€“x(a2+1)

To find the zeros of the equation put g(x) = 0

= a(x2+1)â€“x(a2+1) = 0

= ax2 + a âˆ’ a2x â€“ x = 0

= ax2 âˆ’ a2x â€“ x + a = 0

= ax(x âˆ’ a) âˆ’ 1(x â€“ a) = 0

= (x â€“ a)(ax â€“ 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 = â€“ coefficient of x / coefficient of x2

a + 1/a = â€“ (-(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 â€“ (1 + 2âˆš2)s + âˆš2

Solution:

Given that,

h(s) = 2s2 â€“ (1 + 2âˆš2)s + âˆš2

To find the zeros of the equation put h(s) = 0

= 2s2 â€“ (1 + 2âˆš2)s + âˆš2 = 0

= 2s2 â€“ 2âˆš2s â€“ s + âˆš2 = 0

= 2s(s â€“ âˆš2) -1(s â€“ âˆš2) = 0

= (2s â€“ 1)(s â€“ âˆš2) = 0

x = âˆš2 and x = 1/2

Hence, the zeros of the quadratic equation are âˆš3 and 1.

Now, Verification

As we know that,

Sum of zeros = â€“ coefficient of s / coefficient of s2

âˆš2 + 1/2 = â€“ (-(1 + 2âˆš2)) / 2

(2âˆš2 + 1)/2 = (2âˆš2 +1)/2

Product of roots = constant / coefficient of s2

1/2 x âˆš2 = âˆš2 / 2

âˆš2 / 2 = âˆš2 / 2

Hence the relationship between zeros and their coefficients are verified.

### (x) f(v) = v2 + 4âˆš3v â€“ 15

Solution:

Given that,

f(v) = v2 + 4âˆš3v â€“ 15

To find the zeros of the equation put f(v) = 0

= v2 + 4âˆš3v â€“ 15 = 0

= v2 + 5âˆš3v â€“ âˆš3v â€“ 15 = 0

= v(v + 5âˆš3) â€“ âˆš3 (v + 5âˆš3) = 0

= (v â€“ âˆš3)(v + 5âˆš3) = 0

v = âˆš3 and v = -5âˆš3

Hence, the zeros of the quadratic equation are âˆš3 and -5âˆš3.

Now, for verification

Sum of zeros = â€“ coefficient of v / coefficient of v2

âˆš3 + (-5âˆš3) = â€“ (4âˆš3) / 1

-4âˆš3 = -4âˆš3

Product of roots = constant / coefficient of v2

âˆš3 x (-5âˆš3) = (-15) / 1

-5 x 3 = -15

-15 = -15

Hence the relationship between zeros and their coefficients are verified.

### (xi) p(y) = y2 + (3âˆš5/2)y â€“ 5

Solution:

Given that,

p(y) = y2 + (3âˆš5/2)y â€“ 5

To find the zeros of the equation put f(v) = 0

= y2 + (3âˆš5/2)y â€“ 5 = 0

= y2 â€“ âˆš5/2 y + 2âˆš5y â€“ 5 = 0

= y(y â€“ âˆš5/2) + 2âˆš5 (y â€“ âˆš5/2) = 0

= (y + 2âˆš5)(y â€“ âˆš5/2) = 0

This gives us 2 zeros,

y = âˆš5/2 and y = -2âˆš5

Hence, the zeros of the quadratic equation are âˆš5/2 and -2âˆš5.

Now, Verification

As we know that,

Sum of zeros = â€“ coefficient of y / coefficient of y2

âˆš5/2 + (-2âˆš5) = â€“ (3âˆš5/2) / 1

-3âˆš5/2 = -3âˆš5/2

Product of roots = constant / coefficient of y2

âˆš5/2 x (-2âˆš5) = (-5) / 1

â€“ (âˆš5)2 = -5

-5 = -5

Hence the relationship between zeros and their coefficients are verified.

### (xii) q(y) = 7y2 â€“ (11/3)y â€“ 2/3

Solution:

Given that,

q(y) = 7y2 â€“ (11/3)y â€“ 2/3

To find the zeros of the equation put q(y) = 0

= 7y2 â€“ (11/3)y â€“ 2/3 = 0

= (21y2 â€“ 11y -2)/3 = 0

= 21y2 â€“ 11y â€“ 2 = 0

= 21y2 â€“ 14y + 3y â€“ 2 = 0

= 7y(3y â€“ 2) â€“ 1(3y + 2) = 0

= (3y â€“ 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 = â€“ coefficient of y / coefficient of y2

2/3 + (-1/7) = â€“ (-11/3) / 7

-11/21 = -11/21

Product of roots = constant / coefficient of y2

2/3 x (-1/7) = (-2/3) / 7

â€“ 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 â€“ (-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 â€“ (21/8)x + (5/16)

= x2 â€“ 21/8x + 5/16

To find the zeros we put f(x) = 0

= x2 â€“ 21/8x + 5/16 = 0

= 16x2 â€“ 42x + 5 = 0

= 16x2 â€“ 40x â€“ 2x + 5 = 0

= 8x(2x â€“ 5) â€“ 1(2x â€“ 5) = 0

= (2x â€“ 5) (8x â€“ 1) = 0

= (2x â€“ 5) = 0 and, or (8x â€“ 1) = 0

Hence, the two zeros are 5/2 and 1/8.

### (iii) -2âˆš3, -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 = -2âˆš3 and

Product of zero = -9

Therefore,

The required polynomial f(x) is,

= x2 â€“ (-2âˆš3)x + (-9)

= x2 + 2âˆš3x â€“ 9

To find the zeros we put f(x) = 0

= x2 + 2âˆš3x â€“ 9 = 0

= x2 + 3âˆš3x â€“ âˆš3x â€“ 9 = 0

= x(x + 3âˆš3) â€“ âˆš3(x + 3âˆš3) = 0

= (x + 3âˆš3) (x â€“ âˆš3) = 0

= (x + 3âˆš3) = 0 and, or (x â€“ âˆš3) = 0

Hence, the two zeros are -3âˆš3and âˆš3.

### (iv) -3/2âˆš5, -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 = -3/2âˆš5 and

Product of zero = -1/2

Therefore,

The required polynomial f(x) is,

= x2 â€“ (-3/2âˆš5)x + (-1/2)

= x2 + 3/2âˆš5x â€“ 1/2

To find the zeros we put f(x) = 0

= x2 + 3/2âˆš5x â€“ 1/2 = 0

= 2âˆš5x2 + 3x â€“ âˆš5 = 0

= 2âˆš5x2 + 5x â€“ 2x â€“ âˆš5 = 0

= âˆš5x(2x + âˆš5) â€“ 1(2x + âˆš5) = 0

= (2x + âˆš5) (âˆš5x â€“ 1) = 0

= (2x + âˆš5) = 0 and, or (âˆš5x â€“ 1) = 0

Hence, the two zeros are -âˆš5/2 and 1/âˆš5.

### Question 3. If Î± and Î² are the zeros of the quadratic polynomial f(x) = x2 â€“ 5x + 4, find the value of 1/Î± + 1/Î² â€“ 2Î±Î².

Solution:

Given that,

Î± and Î² 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 = Î±+Î² = -b/a = â€“ (-5)/1 = 5,

Product of the roots = Î±Î² = c/a = 4/1 = 4

We have to find 1/Î± +1/Î² â€“ 2Î±Î²

= [(Î± +Î²)/ Î±Î²] â€“ 2Î±Î²

= (5)/ 4 â€“ 2(4) = 5/4 â€“ 8 = -27/ 4

### Question 4. If Î± and Î² are the zeros of the quadratic polynomial p(y) = 5y2 â€“ 7y + 1, find the value of 1/Î±+1/Î².

Solution:

Given that,

Î± and Î² 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 = Î±+Î² = -b/a = â€“ (-7)/5 = 7/5

Product of the roots = Î±Î² = c/a = 1/5

We have to find 1/Î± +1/Î²

= (Î± +Î²)/ Î±Î²

= (7/5)/ (1/5) = 7

### Question 5. If Î± and Î² are the zeros of the quadratic polynomial f(x)=x2 â€“ x â€“ 4, find the value of 1/Î±+1/Î²â€“Î±Î².

Solution:

Given that,

Î± and Î² are the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = â€“ 4

So, we can find,

Sum of the roots = Î±+Î² = -b/a = â€“ (-1)/1 = 1

Product of the roots = Î±Î² = c/a = -4 /1 = â€“ 4

We have to find, 1/Î± +1/Î² â€“ Î±Î²

= [(Î± +Î²)/ Î±Î²] â€“ Î±Î²

= [(1)/ (-4)] â€“ (-4) = -1/4 + 4 = 15/ 4

### Question 6. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = x2 + x â€“ 2, find the value of 1/Î± â€“ 1/Î².

Solution:

Given that:

Î± and Î² are the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = â€“ 2

So, we can find

Sum of the roots = Î±+Î² = -b/a = â€“ (1)/1 = -1,

Product of the roots = Î±Î² = c/a = -2 /1 = â€“ 2

We have to find, 1/Î± â€“ 1/Î²

= [(Î² â€“ Î±)/ Î±Î²] = [Î²-Î±]/(Î±Î²) x (Î±-Î²)/Î±Î² = (âˆš(Î±+Î²)2 -4Î±Î²) / Î±Î² = âˆš(1+8) / 2 = 3/2

### Question 7. If one of the zero of the quadratic polynomial f(x) = 4x2 â€“ 8kx â€“ 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 = â€“ 9

And, for roots to be negative of each other, let us assume that the roots Î± and â€“ Î±.

Using these values we can find,

Sum of the roots = Î± â€“ Î± = -b/a = â€“ (-8k)/1 = 8k = 0 [âˆµ Î± â€“ Î± = 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 Î± and Î² are the zeros of the quadratic polynomial p(x) = 4x2 â€“ 5x â€“ 1, find the value of Î±2 Î²+Î± Î²2

Solution:

Given that,

Î± and Î² 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 = Î±+Î² = -b/a = â€“ (-5)/4 = 5/4

Product of the roots = Î±Î² = c/a = -1/4

We have to find, Î±^2 Î²+Î± Î²^2

= Î±Î²(Î± +Î²)

= (-1/4)(5/4) = -5/16

### Question 10. If Î± and Î² are the zeros of the quadratic polynomial f(t)=t2â€“ 4t + 3, find the value of Î±4 Î²3+Î±3 Î²4.

Solution:

Given that,

Î± and Î² 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 = Î±+Î² = -b/a = â€“ (-4)/1 = 4 ,

Product of the roots = Î±Î² = c/a = 3/1 = 3

We have to find, Î±4 Î²3 + Î±3 Î²4

= Î±3 Î²3 (Î± +Î²)

= (Î±Î²)3 (Î± +Î²)

= (3)3 (4) = 27 x 4 = 108

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