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
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(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
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A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
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