Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Students can download Maths Chapter 3 Algebra Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Additional Questions

I. Choose the correct answer.

Question 1.
The HCF of x2 – y2; x3 – y3, …………. xn – yn where n ∈ N is
(1) x – y
(2) x + y
(3) xn – yn
(4) do not intersect
Answer:
(1) x – y

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 2.
Which of the following is correct.
(i) Every polynomial has finite number of multiples
(ii) LCM of two polynomials of degree “2” may be a constant
(iii) HCF of 2 polynomials may be a constant
(iv) Degree of HCF of two polynomials is always less than degree of L.C.M.
(1) (i) and (iii)
(2) (iii) and (iv)
(3) (iii) only
(4) (iv) only
Answer:
(3) (iii) only

Question 3.
The HCF of x2 – 2xy + y2 and x4 – y4 is …………….
(1) 1
(2) x + y
(3) x – y
(4) x2 – y2
Answer:
(3) x – y

Question 4.
The L.C.M. of ak ak+3, ak+5 where K ∈ N is …………
(1) ak+5
(2) ak
(3) ak+6
(4) ak+9
Answer:
(1) ak+5

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 5.
The LCM of (x + 1)2 (x – 3) and
(x2 – 9) (x + 1) is
(1) (x + 1)3 (x2 – 9)
(2) (x + 1)2 x2 – 9)
(3) (x + 1)2 (x – 3)
(4) (x – 9) (x + 1)
Answer:
(2) (x + 1)2(x2 – 9)

Question 6.
If \(\frac{a^{3}}{a-b}\) is added with \(\frac{b^{3}}{b-a}\) then the new expressions is …………
(1) a2 – ab + b2
(2) a2 + ab + b2
(3) a3 + b3
(4) a3 – b3
Answer:
(2) a2 + ab + b2

Question 7.
The solution set of x + \(\frac { 1 }{ x } \) = \(\frac { 5 }{ 2 } \) is ………….
(1) 2,\(\frac { 1 }{ 2 } \)
(2) 2,-\(\frac { 1 }{ 2 } \)
(3) -2, – \(\frac { 1 }{ 2 } \)
(4) -2, \(\frac { 7 }{ 2 } \)
Answer:
(1) 2,\(\frac { 1 }{ 2 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 8.
On dividing \(\frac{x^{2}-25}{x+3}\) by \(\frac{x+5}{x^{2}-9}\) is equal to ……………….
(1) (x – 5)(x + 3)
(2) (x + 5) (x – 3)
(3) (x – 5)(x – 3)
(4) (x + 5)(x + 3)
Answer:
(3) (x – 5)(x – 3)

Question 9.
The square root of (x + 11)2 – 44x is ………….
(1)|(x – 11)2
(2) |x + 11|
(3) |11 – x2|
(4) |x – 11|
Answer:
(4) |x – 11|

Question 10.
If α, β are the zeros of the polynomial p(x) = 4x2 + 3x + 7 then \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) is equal to …………
(1) \(\frac { 7 }{ 3 } \)
(2) – \(\frac { 7 }{ 3 } \)
(3) \(\frac { 3 }{ 7 } \)
(4) – \(\frac { 3 }{ 7 } \)
Answer:
(4) – \(\frac { 3 }{ 7 } \)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 11.
The value of Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 1 is  ……….
(1) -5
(2) 5
(3) 4
(4) -3
Answer:
(2) 5

Question 12.
If α and β are the roots of the equation ax2 + bx + c = 0 then (α + β)2 is ……………..
(1) \(\frac{-b^{2}}{a^{2}}\)
(2) \(\frac{-c^{2}}{a^{2}}\)
(3) \(\frac{-b^{2}}{a^{2}}\)
(4) \(\frac { bc }{ a } \)
Answer:
(3) \(\frac{-b^{2}}{a^{2}}\)

Question 13.
The roots of the equation x2 – 8x + 12 = 0 are
(1) real and equal
(2) real and rational
(3) real and irrational
(4) unreal
Answer:
(2) real and rational

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 14.
If one root of the equation is the reciprocal of the other root in ax2 + bx + c = 0 then …………
(1) a = c
(2) a = b
(3) b = c
(4) c = 0
Answer:
(1) a = c

Question 15.
If α and β are the roots of the equation x2 + 2x + 8 = 0 then the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) is ………………
(1) \(\frac { 1 }{ 2 } \)
(2) 6
(3) \(\frac { 3 }{ 2 } \)
(4) –\(\frac { 3 }{ 2 } \)
Answer:
(4) –\(\frac { 3 }{ 2 } \)

Question 16.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 2
are………
(1) 4, 6, 6
(2) 6, 6, 4
(3) 6, 4, 6
(4) 4, 4, 6
Answer:
(3) 6, 4, 6

Question 17.
If [-1 -2 4] Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 3 then the value of “a” is ………….
(1) 2
(2) -4
(3) 4
(4) -2
Answer:
(4) -2

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 18.
The matrix A given by (aij)2×2 if aij = i – j is …………
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 4
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 5

Question 19.
If A is of order 4 × 3 and B is of order 3 × 4 then the order of BA is ………………….
(1) 3 × 4
(2) 4 × 4
(3) 3 × 3
(4) 4 × 1
Answer:
(3) 3 × 3

Question 20.
If Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 6 then “x” is ……………..
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(4) 4

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

II. Answer the following.

Question 1.
Solve x + y = 7; y + z = 4; z + x = 1
Answer:
x + y = 7 ……(1)
y + z = 4 ………(2)
z + x = 1 …………(3)
Adding (1); (2) and (3)
2x + 2y + 2z = 12
x + y + z = 6 ….(4)
From (1) ⇒ x + y = 7
7 + z = 6
z = 6 – 7 = -1
From (2) ⇒ x + 4 = 6
x = 6 – 4 = 2
From (3) ⇒ y + 1 = 6
y = 6 – 1 = 5
The value of x = 2, y = 5 and z = -1

Question 2.
Find the HCF of 25x4y7; 35x3y8; 45x3y3
Answer:
25x4y7 = 5 × 5 × x4 × y7
35x3y8 = 5 × 7 × x3 × y8
45 x3y3 = 3 × 3 × 5 × x3 × y3
H.C.F. = 5x3y3

Question 3.
Find the values of k for which the following equation has equal roots.
(k – 12)x2 + 2(k – 12)x + 2 = 0
Solution:
\(\frac{(k-12)}{a} x^{2}+\frac{2(k-12)}{b} x+\frac{2}{c}=0\)
Δ = b2 – 4ac = (2(k – 12))2 – 4(6 – 12)(2)
= 4(k – 12)[(k – 12) – 2]
= 4(k – 12)(k – 14)
The given equation will have equal roots, if A = 0
⇒ 4(k – 12)(k – 14) = 0
k – 12 = 0 or k – 14 = 0
k = 12, 14

Question 4.
Find the LCM of x3 + y3; x3 – y3; x4 + x2y2 + y4
Answer:
x3 + y3 = (x + y) (x2 – xy + y2)
x3 – y3 = (x – y)(x2 + xy + y2)
x4 + x2y2 + y4 = (x2 + y2)2 – (xy)2
= (x2 + y2 + xy)
L.C.M. = (x + y)(x – y) (x2 + xy + y2)
(x2 – xy + y2)
= [(x + y) (x2 – xy +y2)]
[(x – y) (x2 + xy + y2)]
= (x3 + y3) (x3 – y3)
L.C.M. = x6 – y6

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 5.
The sum of two numbers is 15. If the sum of their reciprocals is \(\frac{3}{10}\), find the numbers.
Solution:
Let the numbers be α, β
Sum of the roots = α + β = 15 ………….. (1)
sum of their reciprocals = \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{10}\) ……….. (2)
\(\frac{\beta+\alpha}{\alpha \beta}=\frac{3}{10}\)
10(α + β) = 3αβ …………. (3)
3αβ = 10 × 15 = 150
Products of the roots = αβ = 50 ………….. (4)
∴ From (1) & (4), we have
x2 – 15x + 50 = 0
(x – 10)(x – 5) = 0 ⇒ x = 10, 5
∴ he numbers are 10, 5.

Question 6.
For What value of k, the G.C.D. of [x2 + x – (2k + 2)] and 2x2 + kx – 12 is (x + 4)?
Answer:
p(x) = x2 + x – (2k + 2)
g(x) = 2x2 + kx – 12
G.C.D. = x + 4
when x + 4 is the G.C.D.
p(-4) = 0 or g(-4) = 0
[Hint: Take any one of the polynomial]
g(x) = 2x2 + kx – 12 = 0
2(-4)2 + k (-4) – 12 = 0
2(16) – 4x – 12 = 0
32 – 4k – 12 = 0
20 = 4k
k = \(\frac { 20 }{ 4 } \) = 5
The value of k = 5

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 7.
Simplify \(\frac{x^{2}+x-6}{x^{2}+4 x+3}\)
Answer:
x2 + x – 6 = (x + 3) (x – 2)
x2 + 4x + 3 = (x + 3) (x + 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 7
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 8

Question 8.
Multiply
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 10
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 11

Question 9.
if P = \(\frac{x^{3}-36}{x^{2}-49}\) and Q = \(\frac { x+6 }{ x+7 } \) find the value of \(\frac { P }{ Q } \).
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 12

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 10.
Simplify
\(\frac { x }{ x+y } \) – \(\frac { y }{ x-y } \)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 13

Question 11.
Find the square root of (x + 11)2 – 44x
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 14

Question 12.
Find the square root of x4 + \(\frac{1}{x^{4}}\) + 2
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 15

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 13.
Solve the equation 2x – 1 – \(\frac { 2 }{ x-2 } \) = 3
Answer:
2x – 1 – \(\frac { 2 }{ x-2 } \) = 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 16
(x – 3) (x – 1) = 0
x – 3 = 0 or x – 1 = 0
x = 3 or x = 1
The solution set is (1,3)

Question 14.
Find the roots of \(\sqrt { 2 }\) x2 + 7x + 5\(\sqrt { 2 }\) = 0
Answer:
\(\sqrt { 2 }\) x2 + 7x + 5 \(\sqrt { 2 }\) = 0
\(\sqrt { 2 }\) x2 + 2x + 5x + 5 \(\sqrt { 2 }\) = 0
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 17
\(\sqrt { 2 }\) x (x + \(\sqrt { 2 }\)) + 5 (x + \(\sqrt { 2 }\)) = 0
(x + \(\sqrt { 2 }\)) (\(\sqrt { 2 }\) x + 5) = 0
(x + \(\sqrt { 2 }\) ) = 0 or \(\sqrt { 2 }\) x + 5 = 0
x = – \(\sqrt { 2 }\) or \(\sqrt { 2 }\) x + 5 = 0
x = – \(\sqrt { 2 }\) or \(\sqrt { 2 }\) x = -5
x = \(\frac{-5}{\sqrt{2}}\)
The roots are and – \(\sqrt { 2 }\) and \(\frac{-5}{\sqrt{2}}\)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 15.
Solve \(\sqrt { x+5 }\) = 2x + 3 using formula method.
Answer:
\(\sqrt { x+5 }\) = 2x + 3
(\(\sqrt { x+5 }\))2 = (2x + 3)2
x + 5 = 4x2 + 9 + 12x
0 = 4x2 + 12x – x + 9 – 5
0 = 4x2 + 11x + 4
Here a = 4, b = 11, c = 4
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 18

Question 16.
The sum of a number and its reciprocal is \(\frac { 37 }{ 6 } \). Find the number.
Answer:
Let the require number be “x”
Its reciprocal is \(\frac { 1 }{ x } \)
By the given data
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 87
The required number is \(\frac { 1 }{ 6 } \) or 6

Question 17.
Determine the nature of the roots of the equation 2x2 + x – 1 = 0
Answer:
2x2 + x – 1 = 0
Here a = 2,b = 1,c = -1
∆ = b2 – 4 ac
= 12 – 4(2) (-1)
= 1 + 8
= 9
Since b2 – 4ac > 0 the roots are real and unequal

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 18.
Find the value of k for which the given equation 9x2 + 3kx + 4 = 0 has real and equal roots.
Answer:
9x2 + 3 kx + 4 = 0
a = 9, b = 5k, c = 4
since the equation has real and equal roots
b2 – 4ac = 0
(3k)2 – 4(9) (4) = 0
9k2 – 144 = 0
9k2 = 144
k2 = \(\frac { 144 }{ 9 } \) = 16
k = \(\sqrt { 16 }\)
k = ± 4

Question 19.
If one root of the equation
3x2 – 10x + 3 = 0 is \(\frac { 1 }{ 3 } \) find the other root
Answer:
α and β are the roots of the equation 3x2 – 10x + 3 = 0
Sum of the roots (α + β) = \(\frac { 10 }{ 3 } \)
Product of the roots (αβ) = \(\frac { 3 }{ 3 } \) = 1
one of the roots is \(\frac { 1 }{ 3 } \) (say α = \(\frac { 1 }{ 3 } \))
αβ = 1
\(\frac { 1 }{ 3 } \) × β = 1
β = 3
The other roots is 3

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 20.
Form the quadratic equation whose roots are 3 + \(\sqrt { 7 }\); 3 – \(\sqrt { 7 }\)
Answer:
Sum of the roots = 3 + \(\sqrt { 7 }\) + 3 – \(\sqrt { 7 }\)
= 6
Product of the roots = (3 + \(\sqrt { 7 }\)) (3 – \(\sqrt { 7 }\) )
= 32 – (\(\sqrt { 7 }\))2
= 9 – 7
= 2
The required equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (6)x + 2 = 0
x – 6x + 2 = 0

Question 21.
If α and β are the roots of the equation 3x2 – 5x + 2 = 0, then find the value of α – β.
Answer:
α and β are the roots of the equation
3x2 – 5x + 2 = 0
α + β = \(\frac { 5 }{ 3 } \), αβ = \(\frac { 2 }{ 3 } \)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 19

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 22.
Determine the matrix A = (aij)3×2 if aij = 3i – 2j
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 20
aij = 3i – 2j
a11 = 3(1) – 2(1) = 3 – 2 = 1
a12 = 3(1) – 2(1) = 3 – 4 = 1
a21 = 3(2) – 2(1) = 6 – 2 = 4
a22 = 3(2) – 2(2) = 6 – 4 = 2
a31 = 3(3) – 2(1) = 9 – 2 = 7
a32 = 3(3) – 2(2) = 9 – 4 = 5
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 21

Question 23.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 22
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 23

Question 24.
Find if
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 24
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 25

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 25.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 26
find BA
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 27

Question 26.
Find the unknowns a, b, c, d, x, y in the given matrix equation.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 28
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 29
Equating the corresponding elements of the two matrices we get
d + 1 = 2
d = 2 – 1 = 1
10 + a = 2a + 1
10 – 1 = 2a – a
9 = a
36 – 2 = b – 5
3b – b = -5 + 2
2b = -3 ⇒ b = \(\frac { -3 }{ 2 } \)
a – 4 = 4c ⇒ a – 4c = 4
9 – 4c = 4 ⇒ 4c = 4 – 9
-4c = -5 ⇒ c = \(\frac { 5 }{ 4 } \)
The value of a = 9, b = \(\frac { -3 }{ 2 } \), c = \(\frac { 5 }{ 4 } \) and d = 1

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 27.
Prove that
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 30
multiplication is inverse to each other.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 31
AB = BA = I
Multiplication of matrices are iverse to each other.

III. Answer the following questions.

Question 1.
Solve x – \(\frac { y }{ 5 } \) = 6; y – \(\frac { z }{ 7 } \) = 8; z – \(\frac { x }{ 2 } \) = 10
Answer:
x – \(\frac { y }{ 5 } \) = 6
multiply by 5
5x – y = 30 …….(1)
y – \(\frac { z }{ 7 } \) = 8
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 32
Substitute the value of x = 8 in (1)
5(8) – y = 30
– y = 30 – 40 = -10
∴ y = 10
Substitute the value of x = 8 in (3)
2z – 8 = 20
2z = 20 + 8
z = \(\frac { 28 }{ 2 } \) = 14
The value of x = 8, y = 10 and z = 14

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 2.
Solve for x,y and z using the given 3 equations
\(\frac { 2 }{ y } \) – \(\frac { 4 }{ z } \) + \(\frac { 3 }{ x } \) = 3; \(\frac { 5 }{ x } \) – \(\frac { 4 }{ y } \) – \(\frac { 8 }{ z } \) = 8 ; \(\frac { 6 }{ y } \) + \(\frac { 6 }{ z } \) +\(\frac { 1 }{ x } \) = 2
Answer:
Let \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
3a + 2b – 4c = 3 ………(1)
5a – 4b – 8c = 8 ………(2)
a + 6b + 6c = 2 ………(3)
(1) × 2 ⇒ 6a + 4b – 8c = 6 …..(1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 33
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 34

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 3.
100 pencils are to be kept inside three types of boxes A, B and C. If 5 boxes of type A, 3 boxes of type B, 2 boxes of type C are used 6 pencils are left out. If 3 boxes of type A, 5 boxes of type B, 2 boxes of type C are used 2 pencils are left out. If 2 boxes of type A, 4 boxes of type B and 4 boxes of type C are used, there is a space for 4 pencils. Find the number of pencils that each box can hold.
Answer:
Let the number of pencil in the box A be “x”
Let the number of pencil in the box B be “y”
Let the number of pencil in the box C be “z”
By the given first condition
5x + 3y + 2z = 94 ….(1)
By the given second condition
3x + 5y + 2z = 98 ….(2)
By the given third condition
2x + 4y + 4z = 104 ….(3)
subtract (1) and (3)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 35
substitute x = 8 and y = 10 in (1)
5(8) + 3(10) + 2z = 94
40 + 30 + 2z = 94
2z = 94 – 70
2z = 24
z = \(\frac { 24 }{ 2 } \) = 12
Number of pencil in box A = 8
Number of pencil in box B = 10
Number of pencil in box C = 12

Question 4.
What 2 masons earn in a day is earned by 3 male workers in a day. The daily wages of 15 female workers is ₹30 more than the total daily wages of 5 masons and 3 male workers. If one mason, one male worker and 2 female workers are engaged for a day, the builder has to pay ?160 as wages. Find the daily wages of a mason, a male worker and a female worker.
Answer:
Let the daily wage of a mason be ₹ x
Let the daily wage of a male worker be ₹ y
Let the daily wage of a female worker be ₹ z
By the given first condition
2x = 3y
2x – 3y = 0 …..(1)
By the given second condition
15z = 5x + 3y + 30
-5x – 3y + 15z = 30
5x + 3y – 15z = -30 ………(2)
By the given third condition
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 36
substitute the value of x = 60 in ….(1)
2(60) – 3y = 0
120 = 3y
y = \(\frac { 120 }{ 3 } \) = 40
substitute the value of x = 60 and y = 40 in (3)
60 + 40 + 2z = 160
2z = 160 – 100
2z = 60
z = \(\frac { 60 }{ 2 } \) = 30
Daily wages of a manson = ₹60
Daily wages of a male worker = ₹40
Daily wages of a female worker = ₹30

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 5.
Find the G.C.D. of x3 – 10x2 + 31x – 30 and 2x3 – 8x2 + 2x + 12
Answer:
p(x) = x3 – 10x2 + 31x – 30
g(x) = 2x3 – 8x2 + 2x + 12
= 2(x3 – 4x2 + x + 6)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 37
G.C.D. = x2 – 5x + 6

Question 6.
The G.C.D of x4 + 3x3 + 5x2 + 26x + 56 and x4 + 2x3 – 4x2 – x + 28 is x2 + 5x + 7. Find their L.C.M.
Answer:
p(x) = x4 + 3x3 + 5x2 + 26x + 56
g(x) = x4 + 2x3 – 4x2 – x + 28
G.C.D. = x2 + 5x + 7
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 38

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 7.
Find the values of “a” and “b” given that p(x) = (x2 + 3x + 2) (x2 – 4x + a); g(x) = (x2 – 6x + 9) × (x2 + 4x + b) and their G.C.D is (x + 2) (x – 3)
Answer:
p(x) = (x2 + 3x + 2) (x2 – 4x + a)
= (x + 1) (x + 2) (x2 – 4x + a)
G.C.D is given as (x + 2) (x – 3)
x – 3 is a factor of x2 – 4x + a
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 39
p(3) = 0
9 – 4(3) + a = 0
9 – 12 + a = 0
– 3 + a =0
a = 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 40
g(x) = (x2 – 6x + 9) (x2 + 4x + 6)
= (x – 3) (x – 3) (x2 + 4x + b)
But G.C.D. is (x + 2) (x – 3)
∴ x + 2 is a factor of x2 + 4x + 6
g(-2) = 0
4 + 4(-2) + b = 0
4 – 8 + 6 = 0
-4 + b = 0
b = 4
The value of a = 3 and b = 4

Question 8.
Find the other polynomial g(x), given that LCM, HCF and p(x) as (x – 1) (x – 2) (x2 – 3x + 3); x – 1 and x3 – 4x2 + 6x – 3 respectively.
Answer:
LC.M. = (x – 1) (x – 2) (x2 – 3x + 3)
HCF = (x – 1)
p(x) = x3 – 4x2 + 6x – 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 41
p(x) = (x – 1) (x2 – 3x + 3)
p(x) × g(x) = LCM × HCF
(x – 1) (x2 – 3x + 3) × g(x)
= (x – 1) (x – 2) (x2 – 3x + 3) (x – 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 42
The other polynominal g(x) x2 – 3x + 2

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 9.
Divide
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 43
Answer:
2x2 + x – 3 = 2x2 + 3x – 2x – 3
= x(2x + 3) – 1 (2x + 3)
= (2x + 3) (x – 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 44
2x2 + 5x + 3 = 2x2 + 3x + 2x + 3
= x(2x + 3) + 1 (2x + 3)
= (2x + 3) (x + 1)
x2 -1 = (x + 1) (x – 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 45
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 46

Question 10.
Simplify
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 47
Answer:
(x2 – x – 6) = (x – 3) (x + 2)
2x2 + 5x – 3 = 2×2 + 6x – x – 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 48
= 2x (x + 3) -1 (x + 3)
= (x + 3) (2x – 1)
x2 + 5x + 6 = (x + 2) (x + 3)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 49
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 50

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 11.
Find the square root of (6x2 + 5x – 6) (6x2 – x – 2) (4x2 + 8x + 3)
Answer:
6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) -2 (2x + 3)
= (2x + 3) (3x – 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 51
6x2 – x – 2 = 6x2 – 4x + 3x – 2
= 2x (3x – 2) + 1 (3x – 2)
= (3x – 2) (2x + 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 52
4x2 + 8x + 3 = 4x2 + 6x + 2x + 3
= 2x(2x + 3) + 1 (2x + 3)
= (2x + 3) (2x + 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 53
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 54

Question 12.
Find the square root of the polynomial
\(\frac{4 x^{2}}{y^{2}}\) + \(\frac { 8x }{ y } \) + 16 + 12 \(\frac { y }{ x } \) + \(\frac{9 y^{2}}{x^{2}}\)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 55
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 555

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 13.
If m – nx + 28x2 + 12x3 + 9x4 is a perfect square, then find the values of m and n.
Answer:
Arrange the polynomial in descending power of x.
9x4 + 12x3 + 28x2 – nx + m
Now,
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 56
Since the given polynomial is a perfect square,
-nx – 16x = 0
-x (n + 16) = 0
n + 16 = 0 ⇒ n = -16
m – 16 = 0 ⇒ m = 16
The value m = 16 and n = -16

Question 14.
If b + \(\frac { a }{ x } \) + \(\frac{13}{x^{2}}\) – \(\frac{6}{x^{3}}\) + \(\frac{1}{x^{4}}\) is a perfect square, find the values of “a” and “b”
Answer:
Arrange the values of “a” and “b”
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 57
Since it is a perfect square
\(\frac { a }{ x } \) + \(\frac { 12 }{ x } \) = 0
\(\frac { 1 }{ x } \) (a + 12) = 0
a + 12 = 0 ⇒ a = -12
b – 4 = 0 ⇒ b = 4
The value of a = -12 and b = 4

Question 15.
Solve
\(\frac { 1 }{ x + 1 } \) + \(\frac { 4 }{ 3x+6 } \) = \(\frac { 2 }{ 3 } \)
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 58
6x2 – 12x + 9x – 18 = 0
6x(x – 2) + 9(x – 2) = 0
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 59
(x – 2) (6x + 9) = 0
x – 2 = 0 or 6x + 9 = 0
x = 2 or 6x + 9 = 0
x = 2 or 6x = -9
x = – \(\frac { 9 }{ 6 } \) = \(\frac { -3 }{ 2 } \)
The solution is \(\frac { -3 }{ 2 } \) or 2

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 16.
A two-digit number is such that the product of the digits is 14. When 45 is added to the number, the digits interchange their places. Find the number (solve by completing square method)
Answer:
Let the ten’s digit be “x”
∴ The unit digit = \(\frac { 14 }{ x } \)
∴ The number is 10x + \(\frac { 14 }{ x } \)
If the digits are interchanged the number is \(\frac { 140 }{ x } \) + x
By the given condition
10x + \(\frac { 14 }{ x } \) + 45 = \(\frac { 140 }{ x } \) + x
multiply by x
10x2 + 14 + 45x = 140 + x2
9x2 + 45x + 14 – 140 = 0
9x2 + 45x – 126 = 0
Divided by 9
x2 + 5x – 14 = 0
x2 + 5x = 14
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 60
Since the digit of the number can not be negative
∴ x = 2
The number = 10x + \(\frac { 14 }{ x } \)
= 20 + \(\frac { 14 }{ 2 } \)
= 20 + 7
= 27
The number is 27

Question 17.
A rectangular garden 10 m by 16 m is to be surrounded by a concreate walk of uniform width. Given that the area of walk is 120 sqm assuming the width of walk be ‘V form the equation then solve it by formula method.
Answer:
Area of the garden = 16 × 10
= 160 sq.m
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 61
Area of the garden with walking area
= (1.6 + 2x) (10 + 2x)
= 160 + 32x + 20x + 4x2
= 4x2 + 52x + 160
Area of the concrete walk = Area of the garden with walk – Area of garden
= 4x2 + 52x + 160 – 160
120 = 4x2 + 52x
4x2 + 52x – 120 = 0
(÷ by 4) ⇒ x2 + 13x – 30 = 0
Here a = 1, b = 13, c = -30
(comparing with ax2 + bx + c = 0)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 62
Since the width cannot be negative. Width of the garden = 2 m

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 18.
If α and β are the roots of the equation 3x2 – 5x + 2 = 0 find the value of
(i) \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)
(ii) α – β
(iii) \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\)
Answer:
Comparing with ax2 + bx + c = 0
a = 3, b = -5, c = 2
α and β are the roots of the equation
3x2 – 5x + 2 = 0
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 63
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 64
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 65
(iii) \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\alpha \beta}\)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 66

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 19.
If α and β are the roots of the equation 3x2 – 6x + 1 = 0 from the equation whose roots are
(i) α2 β;β2α
(ii) 2α + β; 2β + a
Answer:
α and β are the roots of 3x2 – 6x + 1 = 0
α + β = \(\frac { 6 }{ 3 } \) = 2
αβ = \(\frac { 1 }{ 3 } \)
(i) Given the roots are α2β and β2α
Sum of the roots = α2β + β2α
= αβ (α + β)
= \(\frac { 1 }{ 3 } \)(2)
= \(\frac { 2 }{ 3 } \)
Product of the roots = (α2β) x (β2α)
= α2β2
= (αβ)3
= (\(\frac { 1 }{ 3 } \))3
= \(\frac { 1 }{ 27 } \)
The quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (\(\frac { 2 }{ 3 } \)) x + \(\frac { 1 }{ 27 } \) = 0
multiply by 27
27x2 – 18x + 1 = 0

(ii) Given the roots are 2α + β; 2 β + α
Sum of the roots = 2α + β + 2 β + α
= 2(α + β) + (α + β)
= 2(2) + 2
= 6
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 67
The quadratic polynomial is
x2 – (sum of the roots) x + product of the roots = 0
x2 – 6x + \(\frac { 25 }{ 3 } \) = 0
3x2 – 18x + 25 = 0

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 20.
Find X and Y if
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 68Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 69
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 70
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 71
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 72

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 21.
Solve for x,y
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 73
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 74
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 75

Question 22.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 78
Show that A2 – 7A + 1013 = 0
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 79
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 80

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

Question 23.
Verify that (AB)T = BT AT if
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 81
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 82
From (1) and (2) we get
(AB)= BTAT

Question 24.
Draw the graph of y = x2 and hence solve x2 – 4x – 5 = 0.
Answer:
Given equations are y = x2 and x2 – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 83
(ii) Plot the points (- 4,16), (- 3, 9), (- 2,4), (-1, 1), (0,0), (1, 1), (2,4), (3, 9), (4,16), (5,25).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 84
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 85
(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).
The x-coordinates of the points are -1 and 5.
Thus solution set is {- 1, 5}.

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 86

Question 25.
Draw the graph of y = 2x2 + x – 6 and hence solve 2x2 + x – 10 = 0.
Answer:
Given equations are y = x2 and x2 – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 832
(ii) Plot the points (- 4, 22), (- 3, 9), (- 2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations: Subtract 2x2 + x – 10 = 0
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 841
y = 4 is a straight line parallel to X-axis
(v) The straight line and parabola intersect at point (-2.5, 4) and (2, 4).
The x-coordinates of the points are -2.5 and 2.
The solution set is {- 2.5, 2}.
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions 842

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