Pedagogy of Mathematics (Part II) - Algebra, Algebra, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Polynomials, Constants, variables, algebraic expression

1. PEDAGOGY OF
MATHEMATICS – PART II
By
Dr. I. Uma Maheswari
Principal
Peniel Rural College of Education,Vemparali, Dindigul District
iuma_maheswari@yahoo.co.in

2. STD IX
CHAPTER 3 – ALGEBRA
Ex – 3.2

11. Solution:
(i) f(y) = 6y – 3y2 + 3 at
(i) y = 1
f(1) = 6(1) – 3(1)2 + 3
= 6 – 3 + 3
= 6
(ii) y = -1
f(-1) = 6(-1) – 3(-1)2 + 3
= -6 – 3 + 3
= -6
(iii) y = 0
f(0) = 6(0) – 3(0)2 + 3 = 3

12. Solution:
p(x) = x2 – 2√2 x + 1
p(2√2) = (2√2)2 – 2√2 (2√2) + 1
= 4(2) – 4(2) + 1
= 8 – 8 + 1
= 1

13. Solution:
(i) p(x) = x – 3
p(x) = 0
x – 3 = 0
x = 3
(ii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = -5
x = -5 / 2
(iii) q(y) = 2y – 3
q(y) = 0
2y – 3 = 0
2x = 3
x = 3 / 2
(iv) f(z) = 8z
f(z) = 0
8z = 0
z = 0
(iv) p(x) = ax, when a ≠ 0
p(x) = 0
ax = 0
x = 0
(vi) h(x) = ax + b, a ≠ 0, a,b ∈ R
h(x) = 0
ax + b = 0
ax = -b
x = -b / a

14. Solution:
(i) 5x – 6 = 0
5x – 6 = 0
5x = 6
x = 6 / 5
(ii) x + 3 = 0
x + 3 = 0
x = -3
(iii) 10x + 9 = 0
10x + 9 = 0
10x = -9
x = -9 / 10
(iv) 9x – 4 = 0
9x – 4 = 0
9x = 4
x = 4 / 9

15. Solution:
(i) p(x) = 2x – 1, x = 1/2
p(1 / 2) = 2(1 / 2) – 1
= 1 – 1
p(1 / 2) = 0
0 is obtained by applying 1/2, thus, 1 / 2 is
the zero of the polynomial.
(ii) p(x) = x3 – 1, x = 1
p(x) = x3 – 1
p(1) = 13 – 1
p(1) = 0
0 is obtained by applying 1, thus, 1 is the
zero of the polynomial.
iii) p(x) = ax + b, x = -b / a
p(x) = ax + b
p(-b / a) = a(-b / a) + b
p(-b / a) = 0
0 is obtained by applying -b / a, thus, -b / a is the
zero of the polynomial.
(iv) p(x) = (x + 3) (x – 4), x = 4, x = –3
p(x) = (x + 3) (x – 4)
x = 4
p(4) = (4 + 3) (4 – 4)
p(4) = 0
x = -3
p(-3) = (-3 + 3) (-3 – 4)
p(-3) = 0
Thus, 4 and -3 are the zeroes of the polynomial.