# Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x^{2} - 2x - 8 (ii) 4s^{2} - 4s + 1 (iii) 6x^{2} - 3 - 7x

(iv) 4u^{2} + 8u (v) t^{2} - 15 (vi) 3x^{2} - x - 4

**Solution:**

We know that the standard form of the quadratic equation is: ax^{2} + bx + c = 0

Simplify the quadratic polynomial by factorization and find the zeroes of the polynomial.

Now we have to find the relation between the zeroes and the coefficients.

To find the sum of zeroes and the product of zeroes we know that,

Let α and β be the zeros of the polynomial.

Sum of zeroes = - coefficient of x / coefficient of x^{2}

α + β = - b / a

Product of Zeroes = constant term / coefficient of x^{2}

α × β = c / a

Put the values in the above formula and find the relation between the zeroes and the coefficients.

(i) x^{2} - 2x - 8

Let's find the zeros of the polynomial by factorization.

x^{2} - 4x + 2x - 8 = 0

x (x - 4) + 2 (x - 4) = 0

(x - 4) (x + 2) = 0

x = 4 , x = -2 are the zeroes of the polynomial.

Thus, α = 4, β = -2

Now let's find the relationship between the zeroes and the coefficients.

Sum of zeroes = - coefficient of x / coefficient of x^{2}

For x^{2} - 2x - 8,

a = 1, b = - 2, c = - 8

α + β = - b / a

Here,

α + β = - 2 + 4 = 2

- b / a = - (- 2) / 1 = 2

Hence, sum of the zeros α + β = - b/a is verified.

Now, Product of zeroes = constant term / coefficient of x^{2}

α × β = c / a

Here,

α × β = - 2 × 4 = - 8

c / a = - 8 / 1 = - 8

Hence, product of zeros α × β = c / a is verified.

Thus, x = 4, -2 are the zeroes of the polynomial.

(ii) 4s^{2} - 4s + 1

4s^{2} - 2s - 2s + 1 = 0

2s (2s - 1) - 1 (2s - 1) = 0

(2s - 1)(2s - 1) = 0

s = 1/2, s = 1/2 are the zeroes of the polynomial.

Thus, α = 1/2 and β = 1/2

Now, let's find the relationship between the zeroes and the coefficients.

For 4s^{2} - 4s + 1,

a = 4, b = - 4 and c = 1

Sum of zeroes = - coefficient of s / coefficient of s^{2}

α + β = - b/a

Here,

α + β = 1/2 + 1/2 = 1

- b / a = - (- 4) / 4 = 1

Hence, α + β = - b / a, verified

Now, Product of zeroes = constant term / coefficient of s^{2}

α × β = c / a

Here,

α × β = 1/2 × 1/2 = 1/4

c / a = 1/4

Hence, α × β = c / a, verified.

Thus, s = 1/2, 1/2 are the zeroes of the polynomial.

(iii) 6x^{2} - 3 - 7x

6x^{2} - 7x - 3 = 0

6x^{2} - 9x + 2x - 3 = 0

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

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

x = 3/2, x = - 1/3 are the zeroes of the polynomial.

Thus, α = 3/2 and β = -1/3

Now, let's find the relationship between the zeroes and the coefficients:

For 6x^{2} - 3 - 7x,

a = 6, b = - 7 and c = - 3

Sum of zeroes = - coefficient of x / coefficient of x^{2}

α + β = - b / a

Here,

α + β = 3/2 + (-1/3) = 7/6

- b / a = - (-7) / 6 = 7/6

Hence, α + β = - b/a, verified

Now, Product of zeroes = constant term / coefficient of x^{2}

α × β = c / a

Here,

α × β = 3/2 × (- 1/3) = -1/2

c / a = (- 3) / 6 = -1/2

Hence, α × β = c/a, verified.

Thus, x = 3/2, - 1/3 are the zeroes of the polynomial.

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

4u(u + 2) = 0

u = 0, u = - 2 are the zeroes of the polynomial

Thus, α = 0 and β = - 2

Now, let's find the relationship between the zeroes and the coefficients

For 4u^{2} + 8u,

a = 4, b = 8, c = 0

Sum of zeroes = - coefficient of u / coefficient of u^{2}

α + β = - b/a

Here,

α + β = 0 + (- 2) = - 2

- b / a = - (8) / 4 = - 2

Hence, α + β = - b / a, verified

Now, Product of zeroes = constant term / coefficient of u^{2}

α × β = c/a

Here,

α × β = 0 × (- 2) = 0

c / a = 0 / 4 = 0

Hence, α × β = c / a, verified.

Thus, u = 0, - 2 are the zeroes of the polynomial.

(v) t^{2} - 15

t^{2} - 15 = 0

t^{2} = 15

t = ±√15

t = -√15, t = √15 are the zeroes of the polynomial.

Thus, α = -√15 and β = √15

Now, let's find the relationship between the zeroes and the coefficients

For t^{2} - 15,

a = 1, b = 0, c = -15

Sum of zeroes = - coefficient of t / coefficient of t^{2}

α + β = - b / a

Here,

α + β = -√15 + √15 = 0

- b / a = - 0 / 1 = 0

Hence, α + β = - b / a, verified

Now, Product of zeroes = constant term / coefficient of t^{2}

α × β = c / a

Here,

α × β = -√15 × √15 = -15

c / a = -15 / 1 = -15

Hence, α × β = c / a, verified.

Thus, t = -√15, √15 are the zeroes of the polynomial.

(vi) 3x^{2} - x - 4

3x^{2} - x - 4 = 0

3x^{2} - 4x + 3x - 4 = 0

x (3x - 4) + 1(3x - 4) = 0

(x + 1)(3x - 4) = 0

x = - 1, x = 4/3 are the zeroes of the polynomial.

Thus, α = - 1 and β = 4/3

Now, let's find the relationship between the zeroes and the coefficients

For 3x^{2} - x - 4,

a = 3, b = - 1, c = - 4

Sum of zeroes = - coefficient of x / coefficient of x^{2}

α + β = - b / a

Here,

α + β = - 1 + 4/3 = 1/3

- b / a = - (-1) / 3 = 1/3

Hence, α + β = - b / a, verified

Now, Product of zeroes = constant term / coefficient of x^{2}

α × β = c/a

Here,

α × β = - 1 × (4/3) = - 4/3

c / a = - 4/3

Hence, α × β = c / a, verified.

Thus, x = - 1, 4/3 are the zeroes of the polynomial.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 2

**Video Solution:**

## Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x² - 2x - 8 (ii) 4s² - 4s + 1 (iii) 6x² - 3 - 7x (iv) 4u² + 8u (v) t² - 15 (vi) 3x² - x - 4

NCERT Solutions Class 10 Maths Chapter 2 Exercise 2.2 Question 1

**Summary:**

The zeroes of the polynomials (i) x^{2} - 2x - 8 (ii) 4s^{2} - 4s + 1 (iii) 6x^{2} - 3 - 7x (iv) 4u^{2} + 8u (v) t^{2} - 15 (vi) 3x^{2} - x - 4 are i) 4, -2 ii) 1/2,1/2 iii) 3/2, -1/3 iv) 0, -2 v) −√15, √15 and vi) -1, 4/3 respectively and the relationship between the zeroes and the coefficients are verified in each case.

**☛ Related Questions:**

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