Question 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

(i)   x2 – 2x – 8

= x – 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x + 2) (x – 4)

The value of x2 – 2x – 8 is zero if (x + 2) = 0 and (x – 4) = 0

x = -2 or   x = 4

Sum of zeroes = (-2 + 4) = 2 = - coefficient of x

coefficient of x2

Product of zeroes = (-2) × 4 = -8 = Constant term

coefficient of x2

(ii)          4s2 – 4s + 1

= 4s2 – 2s – 2s + 1

= 2s (2s – 1) – 1 (2s – 1)

= ( 2s – 1 )  ( 2s – 1 )

The value of 4s2 – 4s + 1 is zero , if (2s-1) = 0 and (2s-1 ) = 0

s = 1/2 , 1/2

Sum of zeroes = (1/2 + 1/2) = 1   - coefficient of x

coefficient of x2

Product of zeroes =1/2 × 1/2 = 1/4 =  constant term

coefficient of x2

(iii)    6x2 –7x – 3

= 6x – 9x + 2x – 3

= 3x (2x – 3) + 1(2x – 3)

= (3x + 1) (2x – 3)

The value of  6x2 –7x – 3 is zero, if (3x + 1) = 0 and (2x – 3) = 0

X = -1 /3 , 3/2

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

coefficient of x2

Product of zeroes = -1/3 × 3/2 = -3/2 =  constant term

coefficient of x2

(iv)        4u2+8u

4u(u+2)

The value of 4u2+8u is zero, if 4u = 0 and (u+2) =0

u   = 0,  - 2

Sum of zeroes = ( 0+ (-2)) = -2 =  - coefficient of x

coefficient of x2

Product of zeroes = (-2) × 0 = 0 =   constant term

coefficient of x2

(v)

(vi)

3x2–x–4

3x – 4x + 3x – 4

=  x (3x – 4) + 1 (3x – 4)

The value of 3x – x + 4 is zero, if (3x – 4) = 0 and (x + 1) = 0

Sum of zeroes = [4/3 + ( -1)] = 1/3 = - coefficient of x

coefficient of x2

Product of zeroes = (-1) × 4/3 = -4/3 = constant term

coefficient of x2