**Find the zeroes of the following quadraticpolynomials and verify the relationship between the zeroes and thecoefficients.**

**Answer
1** :

**(i) x ^{2}–2x –8**

**Solutions:**

**(i) x ^{2}–2x –8**

x^{2}– 4x+2x–8

= x(x–4)+2(x–4)

= (x-4)(x+2)

Therefore, zeroes of polynomial equation x^{2}–2x–8are (4, -2)

**(ii) 4s ^{2}–4s+1**

**Solutions:**

** (ii) 4s ^{2}–4s+1**

⇒4s^{2}–2s–2s+1

=2s(2s–1)–1(2s-1)

= (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s^{2}–4s+1are (1/2, 1/2)

** **

**(iii) 6x ^{2}–3–7x**

**Solutions:**

** (iii)6x ^{2}–3–7x**

⇒6x^{2}–7x–3

= 6x^{2 }–9x + 2x – 3

= 3x(2x – 3) +1(2x – 3)

= (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x^{2}–3–7xare (-1/3, 3/2)

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

**Solutions:**

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

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u^{2} +8u are (0, -2).

** (v) t ^{2}–15**

**Solutions:**

** (v) t ^{2}–15**

⇒ t^{2} = 15 or t = ±√15

Therefore, zeroes of polynomial equation t^{2} –15are (√15, -√15)

** **

**(vi) 3x ^{2}–x–4**

**Solutions:**

** (vi) 3x ^{2}–x–4**

=3x^{2}–4x+3x–4

= x(3x-4)+1(3x-4)

= (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x^{2} –x – 4 are (4/3, -1)

** **

** **

** **

**Find a quadratic polynomial each with thegiven numbers as the sum and product of its zeroes respectively.**

**Answer
2** :

**(i) 1/4 , -1**

**Solution:**

Let the zeroes of polynomial be α and β.

Now, Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

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

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(1/4)x +(-1) = 0**

**4x ^{2}–x-4 = 0**

**Thus,4x ^{2}–x–4 is the **quadratic polynomial.

** **

** (ii)**√2, 1/3

**Solution:**

Let the zeroes of polynomial be α and β.

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

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

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2} –(**√2

**3x ^{2}-3**√2x+1 = 0

**Thus, 3x ^{2}-3**√2x+1

** **

**(iii) 0, √5**

**Solution:**

Let the zeroes of polynomial be α and β.

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

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

as:-

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(0)x +**√5

**Thus, x ^{2}+**√5

** **

**(iv) 1, 1**

**Solution:**

Let the zeroes of polynomial be α and β.

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

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

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–x+1 = 0**

**Thus , x ^{2}–x+1is the **quadratic polynomial.

** **

**(v) -1/4, 1/4**

**Solution:**

Let the zeroes of polynomial be α and β.

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(-1/4)x +(1/4) = 0**

**4x ^{2}+x+1 = 0**

**Thus,4x ^{2}+x+1 is the **quadratic polynomial.

** **

**(vi) 4, 1**

**Solution:**

Let the zeroes of polynomial be α and β.

Sum of zeroes = α+β =

Product of zeroes = αβ = 1

**x ^{2}–(α+β)x+αβ = 0**

**x ^{2}–4x+1 = 0**

**Thus, x ^{2}–4x+1 is the **quadratic polynomial.

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