Chapter 2: Polynomials Exercise – 2.1

Question: 15

If α and β are the zeroes of the quadratic polynomial f(x) = x² – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.

Solution:

Since, α and β are the zeroes of the quadratic polynomial

f(x) = x² – p(x + 1)– c

Now,

Sum of the zeroes = α + β = p

Product of the zeroes = α × β = (- p – c)

So,

(α + 1)(β + 1)

= αβ + α + β + 1

= αβ + (α + β) + 1

= (− p – c) + p + 1

= 1 – c = RHS

So, LHS = RHS

Hence, proved.

Question: 16

If α and β are the zeroes of the quadratic polynomial such that α + β = 24 and α – β = 8, find a quadratic polynomial having α and β as its zeroes.

Solution:

We have,

α + β = 24         …… E-1

α – β = 8              …. E-2

By solving the above two equations accordingly, we will get

2α = 32 α = 16

Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will get

β = 16 – 8 β = 8

Now,

Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24

Product of the zeroes = αβ = 16 × 8 = 128

Then, the quadratic polynomial is-K 

x²– (sum of the zeroes)x + (product of the zeroes) =  x² – 24x + 128

Hence, the required quadratic polynomial is f(x) = x² + 24x + 128

Question: 17

If α and β are the zeroes of the quadratic polynomial 

f(x) = x² – 1, find a quadratic polynomial whose zeroes are 

Solution:

We have,

f(x) = x² – 1

Sum of the zeroes = α + β = 0

Product of the zeroes = αβ = – 1

From the question,

Sum of the zeroes of the new polynomial

{By substituting the value of the sum and products of the zeroes}

As given in the question,

Product of the zeroes

Hence, the quadratic polynomial is

x² – (sum of the zeroes)x + (product of the zeroes)

= kx² – (−4)x + 4x² –(−4)x + 4 

Hence, the required quadratic polynomial is f(x) = x² + 4x + 4

Question: 18

If α and β are the zeroes of the quadratic polynomial f(x) = x² – 3x – 2, find a quadratic polynomial whose zeroes are

Solution:

We have,

f(x) = x² – 3x – 2

Sum of the zeroes =  α + β = 3

Product of the zeroes = αβ = – 2

From the question,

Sum of the zeroes of the new polynomial

So, the quadratic polynomial is,

x²- (sum of the zeroes)x + (product of the zeroes)

x² - (sum of the zeroes)x + (product of the zeroes)

Hence, the required quadratic polynomial is k 

Question: 19

If α and β are the zeroes of the quadratic polynomial f(x) = x² + px + q, form a polynomial whose zeroes are (α + β)² and (α – β)².

Solution:

We have,

f(x) = x² + px + q

Sum of the zeroes = α + β = -p

Product of the zeroes = αβ = q

From the question,

Sum of the zeroes of new polynomial = (α + β)² + (α – β)²

= (α + β)² + α² + β² – 2αβ

= (α + β)² + (α + β)² – 2αβ – 2αβ

= (- p)² + (- p)² – 2 × q – 2 × q

= p² + p² – 4q

= p² – 4q

Product of the zeroes of new polynomial = (α + β)² (α – β)²

= (- p)²((- p)² - 4q)

= p² (p²–4q)

So, the quadratic polynomial is,

x² – (sum of the zeroes)x + (product of the zeroes)

= x² – (2p² – 4q)x + p²(p² – 4q)

Hence, the required quadratic polynomial is f(x) = k(x² – (2p² –4q) x + p²(p² - 4q)).

Question: 20

If α and β are the zeroes of the quadratic polynomial f(x) = x² – 2x + 3, find a polynomial whose roots are:

(i)  α + 2,β + 2

Solution:

We have,

f(x) = x² – 2x + 3

Sum of the zeroes = α + β = 2

Product of the zeroes = αβ = 3

(i) Sum of the zeroes of new polynomial = (α + 2) + (β + 2)

= α + β + 4

= 2 + 4 = 6

Product of the zeroes of new polynomial = (α + 1)(β + 1)

= αβ + 2α + 2β + 4

= αβ + 2(α + β) + 4 = 3 + 2(2) + 4 = 11

So, quadratic polynomial is:

x² – (sum of the zeroes)x + (product of the zeroes)

= x² – 6x +11

Hence, the required quadratic polynomial is f(x) = k(x² – 6x + 11)

f(x) = k(x² – 6x + 11)

(ii) Sum of the zeroes of new polynomial

Product of the zeroes of new polynomial

So, the quadratic polynomial is,

x² – (sum of the zeroes)x + (product of the zeroes)

Thus, the required quadratic polynomial is f(x) = k(x² – 23x + 13)

Question: 21

If α and β are the zeroes of the quadratic polynomial f(x) = ax² + bx + c, then evaluate:

(i)  α – β

(iv) α²β + αβ²

(v) α⁴ + β⁴

Solution:

f(x) = ax² + bx + c

Here,

Sum of the zeroes of polynomial = α + β = -b/a

Product of zeroes of polynomial = αβ = c/a

Since, α + β are the roots (or) zeroes of the given polynomial, so

(i) α – β

The two zeroes of the polynomials are -

From previous question we know that,

Also,

αβ = c/a

Putting the values in E.1, we will get

Since,

Sum of the zeroes of polynomial = α + β = – b/a

Product of zeroes of polynomial = αβ = c/a

After substituting it in E-1, we will get

(iv) α²β + αβ²

= αβ(α + β) …….. E- 1.

Since,

Sum of the zeroes of polynomial = α + β = – b/ a

Product of zeroes of polynomial = αβ = c/a

After substituting it in E-1, we will get

(v)  α⁴ + β⁴

= (α² + β²)² – 2α²β²

= ((α + β)² – 2αβ)² – (2αβ)² ……. E- 1

Since,

Sum of the zeroes of polynomial = α + β = – b/a

Product of zeroes of polynomial = αβ = c/a

After substituting it in E-1, we will get

Since,

Sum of the zeroes of polynomial = α + β = – b/a

Product of zeroes of polynomial = αβ = c/a

After substituting it, we will get

Since,

Sum of the zeroes of polynomial = α + β = – b/a

Product of zeroes of polynomial = αβ = c/a

After substituting it, we will get

Since,

Sum of the zeroes of polynomial= α + β = – b/a

Product of zeroes of polynomial= αβ = c/a

After substituting it, we will get