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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

Solution:

We know that the standard form of the quadratic equation is: ax2 + 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 x2

α + β = - b / a

Product of Zeroes = constant term / coefficient of x2

α × β = c / a

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

(i) x2 - 2x - 8

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

x2 - 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 x2

For x2 - 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 x2

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) 4s2 - 4s + 1

4s2 - 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 4s2 - 4s + 1,

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

Sum of zeroes = - coefficient of s / coefficient of s2

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 s2

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) 6x2 - 3 - 7x

6x2 - 7x - 3 = 0

6x2 - 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 6x2 - 3 - 7x,

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

Sum of zeroes = - coefficient of x / coefficient of x2

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 x2

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) 4u2 + 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 4u2 + 8u,

a = 4, b = 8, c = 0

Sum of zeroes = - coefficient of u / coefficient of u2

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 u2

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) t2 - 15

t2 - 15 = 0

t2 = 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 t2 - 15,

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

Sum of zeroes = - coefficient of t / coefficient of t2

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 t2

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) 3x2 - x - 4

3x2 - x - 4 = 0

3x2 - 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 3x2 - x - 4,

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

Sum of zeroes = - coefficient of x / coefficient of x2

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 x2

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) x2 - 2x - 8 (ii) 4s2 - 4s + 1 (iii) 6x2 - 3 - 7x (iv) 4u2 + 8u (v) t2 - 15 (vi) 3x2 - 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.

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