**Q.1** **Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.**

**(i) 6x² + 11x + 5 = 0**

**Sol _{n}: –**

**6x ^{2 }+ 11x + 5 **= 6x

^{2 }+ 6x + 5x + 5

= 6x(x +1) + 5(x +1)

= (x +1) (6x +5)

∴ zeroes of polynomial equation** 6x ^{2} +11x +5 **are { −1,

Now, **Sum of zeroes** of this given polynomial equation = −1+( **= −116**

But, the Sum of zeroes of any quadratic polynomial equation is given by =

=** −116**

And Product of these zeroes will be = **= 56**

But, the Product of zeroes of any quadratic polynomial equation is given by =

=

**Hence the relationship is verified**.

**(ii) 4s ^{2} – 4s + 1**

**Sol:**

** 4s ^{2} – 4s + 1 = **4s

^{2}– 2s

**–**2s + 1

= 2s (2s **– **1) **–**1(2s **– **1)

= (2s **– **1) (2s **– **1)

∴ zeroes of the given polynomial are: **{ 12,12}**

∴ Sum of these zeroes will be = = **1.**

But, The Sum of zeroes of any quadratic polynomial equation is given by =

=

And the Product of these zeroes will be =

=

But, Product of zeroes in any quadratic polynomial equation is given by =

=

**Hence, the relationship is verified.**

**(iii) 6x ^{2} – 3 – 7x**

**Sol:**

** 6x ^{2} – 7x – 3 = **6x

^{2}– 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3)

= (3x + 1) (2x – 3)

∴ zeroes of the given polynomial are: – (

∴ sum of these zeroes will be =

=

But, The Sum of zeroes in any quadratic polynomial equation is given by =

=

And Product of these zeroes will be =

Also, the Product of zeroes in any quadratic polynomial equation is given by =

=

**Hence, the relationship is verified.**

**(iv) 4u ^{2} + 8u**

**Sol:**

** 4u ^{2} + 8u = **4u (u+2)

Clearly, for finding the zeroes of the above quadratic polynomial equation either: – **4u=0** or **u+2=0**

Hence, the zeroes of the above polynomial equation will be (**0, −2)**

∴ Sum of these zeroes will be = **−2**

But, the Sum of the zeroes in any quadratic polynomial equation is given by =

** = −84 = −2**

And product of these zeroes will be = 0 × −2 = **0**

But, the product of zeroes in any quadratic polynomial equation is given by = ** −04 = 0**

**Hence, the relationship is verified.**

**(v) t ^{2} – 15**

**Sol:**

** t ^{2} – 15 **= (t+

Therefore, zeroes of the given polynomial are: – **{ 15−−√, −15−−√}**

∴ sum of these zeroes will be = **0**

But, the Sum of zeroes in any quadratic polynomial equation is given by =

** = −01 = 0 **

And the product of these zeroes will be = (**−15**

But, the product of zeroes in any quadratic polynomial equation is given by

= ** = −151 = −15**

**Hence, the relationship is verified.**

**(vi) 3x ^{2} – x – 4**

**Sol:**

** 3x ^{2} − x − 4 = **3x

^{2}– 4x + 3x − 4

= x (3x – 4) +1(3x – 4)

= ( x + 1) (3x – 4)

∴ zeroes of the given polynomial are: – **{−1, 43 }**

∴ sum of these zeroes will be = −1 +

But, the Sum of zeroes in any quadratic polynomial equation is given by =

=

And the Product of these zeroes will be = {−1 ×

=

But, the Product of zeroes in any quadratic polynomial equation is given by =

**= −43**

**Hence, the relationship is verified.**

**Q2. Form a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

**(i). 26 , −3**

**Sol. **

Given,

α + β =** 26**

αβ = **−3**

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2} – (α+β)x +αβ = 0**

Thus, the required quadratic equation will be:

x^{2} – (

**6x ^{2 }− 2x – 18 = 0.**

**(ii). 3–√ , 43**

**Sol. **

Given,

α + β =

αβ =

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

**x ^{2} – (α+β)x +αβ=0**

Thus, the required quadratic equation will be: –

x^{2} – (

**3x ^{2 }− 33–√x + 4 = 0.**

**(iii).** ** 0, 7–√**

**Sol. **

Given,

α + β = **0**

αβ =

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2} – (α+β)x +αβ=0**

Thus, the required quadratic equation will be: –

x^{2} – (0)x +

**x ^{2 }+ 7–√ = 0.**

**(iv). −2, −2**

Sol.

Given,

α + β = **−2**

αβ = **−2**

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

**x ^{2} – (α+β)x +αβ=0**

∴ The required quadratic polynomial will be:

x^{2} – (−2)x −2 = 0

**x ^{2 }+ 2x – 2 = 0.**

**(v).**

Sol.

Given,

α + β =

αβ =

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

**x ^{2} – (α+β)x +αβ = 0**

∴ The required quadratic polynomial will be:

x^{2} – (

**18x ^{2 }+ 63x + 6 = 0.**

**(vi).** **6, 0**

Sol.

Given,

α + β = **6**

αβ = **0**

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2} – (α+β)x +αβ = 0**

∴ The required quadratic polynomial will be:-

x^{2} – 6x + 0 = 0

**x ^{2} – 6x = 0. **

** EXTRA QUESTIONS**

**Q.1 Find a quadratic polynomial whose zeroes are: – 2+12√, 2−12√.**

**Sol. **

Given: –

α + β =

=** 4**.

αβ =

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

**x ^{2} – (α+β)x +αβ = 0**

Thus, the required quadratic polynomial equation will be :-

x^{2} – (4)x −

**2x ^{2}−8x+7 = 0.**

**Q.2** **If α and β are the roots of a quadratic polynomial ax ^{2}+bx+c, then find the value of α^{2} + β^{2}.**

**Sol.**

From the equation,

And,

∴

∴

∴

** ∴ α2+β2=b2−2aca2.**

Similarly, **we can find out the values of (α ^{3} + β^{3}) and (α^{3} – β^{3}).**

**Q3. If α and β are zeroes of a quadratic polynomial x ^{2}+4x+3, form the polynomial whose zeroes are 1+αβand1+βα.**

**Sol.**

Since α and β are zeroes of a quadratic polynomial x^{2}+4x+3,

α+β= **-4**, αβ =** 3**

Given: – α_{1} =

β_{1 }=

Now, sum of zeroes =

=

=

**=**

On putting values of α+β and αβ from above we get:-

Sum of zeroes = **α _{1} + β_{1}** =

**=**

Now, Product of zeroes =

=

=

**= (α+β)2αβ**

On putting values of α+β and αβ we get:

Product of zeroes = **α _{1} × β_{1}** =

**= 163**

Thus the required quadratic polynomial equation will be:-

x^{2} – (α_{1} + β_{1})x + α_{1}β_{1} = 0

x^{2} – (

**3x ^{2} – 16x +16=0.**

**Q4. If α and β are zeroes of a quadratic polynomial p(x) = rx ^{2}+4x+4, Find the values of “r” if: – α^{2} + β^{2} = 24.**

Sol.

From the given polynomial p(x),

α + β **= −4r**, and αβ =

**………. (1)**−4r

Since,**(α + β) ^{2} = α^{2} + β^{2} + 2αβ**

Given, **α ^{2} + β^{2} = 24** and from

**equation (1).**

Therefore,

16 = 24 r^{2 }+ 8r

3r^{2 }+ r – 2 = 0

3r^{2 }+ 3r – 2r -2 = 0

3r(r+1) – 2(r+1) = 0

(3r-2) (r+1) = 0

**∴ r = 23** or

**r = -1**

** ****DIVISION ALOGORITHM**

If suppose p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that: –

**p(x)=g(x) × q(x) + r(x)**, where r(x)=0 or degree of r(x) < degree of g(x) …….(1)

This is known as the **division algorithm for polynomials.**

And since**, Dividend = Divisor × Quotient + Remainder**

On comparing it with equation (1) we can conclude that:

**Dividend = p(x).**

**Divisor = g(x).**

**Quotient = q(x).**

**Remainder = r(x).**