Question

give me solutions of ex 2.2 for class 10

Answer 1

. 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

Answer

(i) x2 – 2x – 8

= (x - 4) (x + 2)

The value of x2 – 2x – 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., whenx = 4 or x = -2

Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.

Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient ofx2

Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient ofx2

(ii) 4s2 – 4s + 1

= (2s-1)2

The value of 4s2 - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient ofs)/Coefficient of s2

Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s2.

(iii) 6x2 – 3 – 7x

= 6x2 – 7x – 3

= (3x + 1) (2x - 3)

The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2

Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient ofx)/Coefficient of x2

Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.

(iv) 4u2 + 8u

= 4u2 + 8u + 0

= 4u(u + 2)

The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2

Therefore, the zeroes of 4u2 + 8u are 0 and - 2.

Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient ofu2

Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient ofu2.

(v) t2 – 15

= t2 - 0.t - 15

= (t - √15) (t + √15)

The value of t2 - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15

Therefore, the zeros of t2 - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient oft)/Coefficient of t2

Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.

(vi) 3x2 – x – 4

= (3x - 4) (x + 1)

The value of 3x2 – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1

Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient ofx)/Coefficient of x2

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.

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

(i) 1/4 , -1

(ii) √2 , 1/3

(iii) 0, √5

(iv) 1,1

(v) -1/4 ,1/4

(vi) 4,1

Answer

(i) 1/4 , -1

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß

α + ß = 1/4 = -b/a

αß = -1 = -4/4 = c/a

If a = 4, then b = -1, c = -4

Therefore, the quadratic polynomial is 4x2 - x -4.

(ii) √2 , 1/3

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß

α + ß = √2 = 3√2/3 = -b/a

αß = 1/3 = c/a

If a = 3, then b = -3√2, c = 1

Therefore, the quadratic polynomial is 3x2 -3√2x +1.

(iii) 0, √5

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß

α + ß = 0 = 0/1 = -b/a

αß = √5 = √5/1 = c/a

If a = 1, then b = 0, c = √5

Therefore, the quadratic polynomial is x2 + √5.

(iv) 1, 1

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß

α + ß = 1 = 1/1 = -b/a

αß = 1 = 1/1 = c/a

If a = 1, then b = -1, c = 1

Therefore, the quadratic polynomial is x2 - x +1.

(v) -1/4 ,1/4

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß

α + ß = -1/4 = -b/a

αß = 1/4 = c/a

If a = 4, then b = 1, c = 1

Therefore, the quadratic polynomial is 4x2 + x +1.

(vi) 4,1

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß

α + ß = 4 = 4/1 = -b/a

αß = 1 = 1/1 = c/a

If a = 1, then b = -4, c = 1

Therefore, the quadratic polynomial is x2 - 4x +1.

Answer 2

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