Exercise 2.1
1The graphs of y = p(x) are given in Fig. 2.10 for some polynomials p(x). Find the number of zeroes of p(x), in each case.
The number of zeroes of p(x) is the number of points where the graph of y = p(x) meets the x-axis.
(i)
The graph is parallel to the x-axis. It does not meet the x-axis at all.
Number of zeroes 0
(ii)
The graph meets the x-axis at only one point.
Number of zeroes 1
(iii)
The graph meets the x-axis at three points.
Number of zeroes 3
(iv)
The graph meets the x-axis at two points.
Number of zeroes 2
(v)
The graph meets the x-axis at four points.
Number of zeroes 4
(vi)
The graph meets the x-axis at three points.
Number of zeroes 3
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Exercise 2.2
1Find 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
For ax2 + bx + c: sum of zeroes = −ba and product of zeroes = ca.
(i) x2 – 2x – 8
x2 – 2x – 8 = x2 – 4x + 2x – 8 = (x – 4)(x + 2)
Zeroes 4 and –2
Sum of zeroes = 4 + (–2) = 2 = –(–2)1 = –(Coefficient of x)Coefficient of x2
Product of zeroes = 4 × (–2) = –8 = –81 = Constant termCoefficient of x2
(ii) 4s2 – 4s + 1
4s2 – 4s + 1 = (2s – 1)2 = (2s – 1)(2s – 1)
Zeroes 12 and 12
Sum of zeroes = 12 + 12 = 1 = –(–4)4 = –(Coefficient of s)Coefficient of s2
Product of zeroes = 12 × 12 = 14 = Constant termCoefficient of s2
(iii) 6x2 – 3 – 7x
6x2 – 7x – 3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (3x + 1)(2x – 3)
Zeroes 32 and –13
Sum of zeroes = 32 + (–13) = 9 – 26 = 76 = –(–7)6 = –(Coefficient of x)Coefficient of x2
Product of zeroes = 32 × (–13) = –12 = –36 = Constant termCoefficient of x2
(iv) 4u2 + 8u
Zeroes 0 and –2
Sum of zeroes = 0 + (–2) = –2 = –84 = –(Coefficient of u)Coefficient of u2
Product of zeroes = 0 × (–2) = 0 = 04 = Constant termCoefficient of u2
(v) t2 – 15
t2 – 15 = (t – √15)(t + √15)
Zeroes √15 and –√15
Sum of zeroes = √15 + (–√15) = 0 = –01 = –(Coefficient of t)Coefficient of t2
Product of zeroes = √15 × (–√15) = –15 = –151 = Constant termCoefficient of t2
(vi) 3x2 – x – 4
3x2 – x – 4 = 3x2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Zeroes 43 and –1
Sum of zeroes = 43 + (–1) = 13 = –(–1)3 = –(Coefficient of x)Coefficient of x2
Product of zeroes = 43 × (–1) = –43 = Constant termCoefficient of x2
2Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) 14, –1 (ii) √2, 13 (iii) 0, √5 (iv) 1, 1 (v) –14, 14 (vi) 4, 1
A quadratic polynomial with sum of zeroes S and product P is k[x2 – Sx + P], where k is any non-zero constant.
(i) Sum = 14, Product = –1
x2 – 14x – 1 ; taking k = 4 : 4x2 – x – 4
Required polynomial 4x2 – x – 4
(ii) Sum = √2, Product = 13
x2 – √2x + 13 ; taking k = 3 : 3x2 – 3√2x + 1
Required polynomial 3x2 – 3√2x + 1
(iii) Sum = 0, Product = √5
Required polynomial x2 + √5
(iv) Sum = 1, Product = 1
Required polynomial x2 – x + 1
(v) Sum = –14, Product = 14
x2 – (–14)x + 14 = x2 + 14x + 14 ; taking k = 4 : 4x2 + x + 1
Required polynomial 4x2 + x + 1
(vi) Sum = 4, Product = 1
Required polynomial x2 – 4x + 1
