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**Question: 1**

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:

(i) f(x) = x^{2} – 2x – 8

(ii) g(s) = 4s^{2} – 4s + 1

(iii) 6x^{2} – 3 – 7x

(iv) h(t) = t^{2} – 15

(v) p(x) = x^{2} + 2√2 x – 6

(vi) q(x) = √3 x^{2} + 10x + 7√3

(vii) f(x) = x^{2} - (√3 + 1)x + √3

(viii) g(x) = a(x^{2} + 1) – x(a^{2} + 1)**Solution: ****(i) f(x) = x**^{2} **– 2x – 8**

We have,

f(x) = x^{2} – 2x – 8

= x^{2} – 4x + 2x – 8

= x (x – 4) + 2(x – 4)

= (x + 2)(x – 4)

Zeroes of the polynomials are – 2 and 4.

Now,

Hence, the relationship is verified.

**(ii) g(s) = 4s**^{2}** – 4s + 1**

We have,

g(s) = 4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

= 2s(2s – 1)− 1(2s – 1)

= (2s – 1)(2s – 1)

Zeroes of the polynomials are 1/2 and 1/2.

Hence, the relationship is verified.

**(iii) 6s**^{2} **− 3 − 7x**

= 6s^{2} − 7x − 3 = (3x + 11) (2x – 3)

Zeros of the polynomials are 3/2 and (-1)/3

Hence, the relationship is verified.

**(iv) h(t) = t**^{2}** – 15**

We have,

Zeroes of the polynomials are - √15 and √15

Sum of the zeroes = 0 - √15 + √15 = 0

0 = 0

Hence, the relationship verified.

**(v)** **p(x) = x ^{2} + 2√2 x – 6**

We have,

p(x) = x

= x

= x(x + 3√2) - √2(x + 3√2)

= (x + 3√2)(x - √2)

Zeroes of the polynomials are 3√2 and –3√2 Sum of the zeroes

- 6 = – 6

Hence, the relationship is verified.

**(vi) q(x) = √3 x ^{2} + 10x + 7√3**

q(x) = √3x^{2} + 10x + 7√3

= √3x^{2} + 7x + 3x + 7√3

= √3x(x+√3)7(x+√3)

= (x + √3)(7 + √3x)

Zeros of the polynomials are -√3 and -7/√3

Product of the polynomials are - √3, 7/√3

7 = 7

Hence, the relationship is verified.

Zeros of the polynomials are 1 and √3

Hence, the relationship is verified

**(viii) g(x) = a[(x**^{2}** + 1)– x(a ^{2} + 1)]2**

= ax

= ax

= ax

= ax(x − a) − 1(x – a) = (x – a)(ax – 1)

Zeros of the polynomials are 1/a and 1 Sum of the zeros

Product of zeros = a/a

Hence, the relationship is verified.

**Question: 2****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} – 5x + 4, find the value of **

We have,

α and β are the roots of the quadratic polynomial.

f(x) = x

Sum of the roots = α + β = 5

Product of the roots = αβ = 4

So,

**Question: 3****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} – 5x + 4, find the value of **

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

p(y) = x^{2} – 5x + 4

Sum of the zeroes = α + β = 5

Product of the roots = αβ = 4

So,

**Question: 4****If α and β are the zeroes of the quadratic polynomial p (y) = 5y ^{2} – 7y + 1, find the value of**

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

p(y) = 5y^{2} – 7y + 1

Sum of the zeroes = α + β = 7

Product of the roots = αβ = 1

So,

**Question: 5****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} – x – 4, find the value of **

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

We have,

f(x) = x^{2} – x – 4

Sum of zeroes = α + β = 1

Product of the zeroes = αβ = - 4

So,

**Question: 6****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} + x – 2, find the value of **

**Solution:**

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

We have,

f(x) = x^{2} + x – 2

Sum of zeroes = α + β = 1

Product of the zeroes = αβ = – 2

So,

**Question: 7****If one of the zero of the quadratic polynomial f(x) = 4x ^{2} – 8kx – 9 is negative of the other, then find the value of k.Solution:**

Let, the two zeroes of the polynomial f(x) = 4x^{2} – 8kx – 9 be α and − α.

Product of the zeroes = α × − α = – 9

Sum of the zeroes = α + (− α) = – 8k = 0

Since, α – α = 0

⇒ 8k = 0 ⇒ k = 0

**Question: 8****If the sum of the zeroes of the quadratic polynomial f(t) = kt ^{2} + 2t + 3k is equal to their product, then find the value of k.**

Let the two zeroes of the polynomial f(t) = kt^{2} + 2t + 3k be α and β.

Sum of the zeroes = α + β = 2

Product of the zeroes = α × β = 3k

Now,

**Question: 9****If α and β are the zeroes of the quadratic polynomial p(x) = 4x ^{2} – 5x – 1, find the value of α^{2}β + αβ^{2}.**

**Solution: **

Since, α and β are the zeroes of the quadratic polynomial p(x) = 4x^{2} – 5x – 1

So, Sum of the zeroes α + β = 5/4

Product of the zeroes α × β = – ¼

Now,

α^{2}β + αβ^{2} = αβ (α + β)

**Question: 10****If α and β are the zeroes of the quadratic polynomial f(t) = t ^{2} – 4t + 3, find the value of α^{4}β^{3} + α^{3}β^{4}.**

Since, α and β are the zeroes of the quadratic polynomial f(t) = t^{2} – 4t + 3

So, Sum of the zeroes = α + β = 4

Product of the zeroes = α × β = 3

Now,

α^{4}β^{3} + α^{3}β^{4} = α^{3}β^{3}(α + β)

= (3)^{3}(4) = 108

**Question: 11****If α and β are the zeroes of the quadratic polynomial f(x) = 6x ^{2} + x – 2, find the value of**

Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x^{2} + x – 2.

Sum of the zeroes = α + β = -⅙

Product of the zeroes =α × β = -⅓

Now,

By substitution the values of the sum of zeroes and products of the zeroes, we will get

= - 25/12

**Question: 12****If α and β are the zeroes of the quadratic polynomial f(x) = 6x ^{2} + x – 2, find the value of **

Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x^{2} + x – 2.

Sum of the zeroes = α + β = 6/3

Product of the zeroes = α × β = 4/3

Now,

By substituting the values of sum and product of the zeroes, we will get

**Question: 13****If the squared difference of the zeroes of the quadratic polynomial f(x) = x ^{2} + px + 45 is equal to 144, find the value of p.Solution**

Let the two zeroes of the polynomial be αand β.

We have,

f(x) = x

Now,

Sum of the zeroes = α + β = – p

Product of the zeroes = α × β = 45

So,

Thus, in the given equation, p will be either 18 or -18.

**Question: 14****If α and β are the zeroes of the quadratic polynomialf(x) = x ^{2} – px + q, prove that **

**Solution:**

Since, α and β are the roots of the quadratic polynomial given in the question.

f(x) = x^{2} – px + q

Now,

Sum of the zeroes = p = α + β

Product of the zeroes = q = α × β

LHS = RHS

Hence, proved.

**Question: 15****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.**

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

f(x) = x^{2} – 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.**

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) = x^{2} – 24x + 128

Hence, the required quadratic polynomial is f(x) = x^{2} + 24x + 128

**Question: 17****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} – 1, find a quadratic polynomial whose zeroes are **

We have,

f(x) = x^{2} – 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

x^{2} – (sum of the zeroes)x + (product of the zeroes)

= kx^{2} – (−4)x + 4x^{2} –(−4)x + 4

Hence, the required quadratic polynomial is f(x) = x^{2} + 4x + 4

**Question: 18****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} – 3x – 2, find a quadratic polynomial whose zeroes are**

We have,

f(x) = x

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,

x

x

Hence, the required quadratic polynomial is k

Hence, the required quadratic polynomial is k

**Question: 19****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} + px + q, form a polynomial whose zeroes are (α + β)^{2} and (α – β)^{2}.**

We have,

f(x) = x

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)

= (α + β)

= (α + β)

= (- p)

= p

= p

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

= (- p)

= p

So, the quadratic polynomial is,

x

= x

Hence, the required quadratic polynomial is f(x) = k(x

**Question: 20****If α and β are the zeroes of the quadratic polynomial f(x) = x ^{2} – 2x + 3, find a polynomial whose roots are:(i) α + 2,β + 2(ii) **

We have,

f(x) = x

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:

x^{2} – (sum of the zeroes)x + (product of the zeroes)

= x^{2} – 6x +11

Hence, the required quadratic polynomial is f(x) = k(x^{2} – 6x + 11)

f(x) = k(x^{2} – 6x + 11)

(ii) Sum of the zeroes of new polynomial

Product of the zeroes of new polynomial

So, the quadratic polynomial is,

x^{2} – (sum of the zeroes)x + (product of the zeroes)

Thus, the required quadratic polynomial is f(x) = k(x^{2} – 23x + 13)

**Question: 21****If α and β are the zeroes of the quadratic polynomial f(x) = ax ^{2} + bx + c, then evaluate:**

(iv) α

(v) α

f(x) = ax^{2} + 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 the 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) α^{2}β + αβ^{2}

= 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) α^{4} + β^{4}

= (α^{2} + β^{2})^{2} – 2α^{2}β^{2}

= ((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

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