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

**Solutions:****(i) x ^{2}–2x –8**

Therefore, zeroes of polynomial equation x

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x

**(ii) 4s ^{2}–4s+1**

⇒ 4s

Therefore, zeroes of polynomial equation 4s

Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s

**(iii) 6x ^{2}–3–7x**

⇒ 6x

Therefore, zeroes of polynomial equation 6x

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x

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

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u

**(v) t ^{2}–15**

⇒ t

Therefore, zeroes of polynomial equation t

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t

**(vi) 3x ^{2}–x–4**

⇒ 3x

Therefore, zeroes of polynomial equation3x

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

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

**(ii)**√2, 1/3**Solution:**

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ 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**

**(iii) 0, √5****Solution:**

Given,

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

∴ 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**

**(iv) 1, 1****Solution:**

Given,

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

∴ 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**

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

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

∴ 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**

**(vi) 4, 1****Solution:**

Given,

Sum of zeroes = α+β =

Product of zeroes = αβ = 1

∴ 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**

**Check out the NCERT Solutions of all the exercises of Polynomials: **

Exercise 2.1. NCERT Solutions: Polynomials

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