Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Additional Questions

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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

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 a^k a^k+3, a^k+5 where K ∈ N is …………
(1) a^k+5
(2) a^k
(3) a^k+6
(4) a^k+9
Answer:
(1) a^k+5

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,\(\frac { 1 }{ 2 } \)

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) – \(\frac { 3 }{ 7 } \)

Question 11.
The value of 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

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.

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] then the value of “a” is ………….
(1) 2
(2) -4
(3) 4
(4) -2
Answer:
(4) -2

Question 18.
The matrix A given by (a_ij)_2×2 if a_ij = i – j is …………

Answer:

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 then “x” is ……………..
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(4) 4

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

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

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)

Question 8.
Multiply

Answer:

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:

Question 10.
Simplify
\(\frac { x }{ x+y } \) – \(\frac { y }{ x-y } \)
Answer:

Question 11.
Find the square root of (x + 11)² – 44x
Answer:

Question 12.
Find the square root of x⁴ + \(\frac{1}{x^{4}}\) + 2
Answer:

Question 13.
Solve the equation 2x – 1 – \(\frac { 2 }{ x-2 } \) = 3
Answer:
2x – 1 – \(\frac { 2 }{ x-2 } \) = 3

(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

\(\sqrt { 2 }\) 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 \(\frac{-5}{\sqrt{2}}\)

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

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

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

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

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 = \(\frac { 2 }{ 3 } \)

Question 22.
Determine the matrix A = (a_ij)_3×2 if a_ij = 3i – 2j
Answer:

a_ij = 3_i – 2_j
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

Question 23.

Answer:

Question 24.
Find if

Answer:

Question 25.

find BA
Answer:

Question 26.
Find the unknowns a, b, c, d, x, y in the given matrix equation.

Answer:

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

Question 27.
Prove that

multiplication is inverse to each other.
Answer:

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

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

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)

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)

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

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

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)

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

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

p(3) = 0
9 – 4(3) + a = 0
9 – 12 + a = 0
– 3 + a =0
a = 3

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

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)

The other polynominal g(x) x² – 3x + 2

Question 9.
Divide

Answer:
2x² + x – 3 = 2x² + 3x – 2x – 3
= x(2x + 3) – 1 (2x + 3)
= (2x + 3) (x – 1)

2x² + 5x + 3 = 2x² + 3x + 2x + 3
= x(2x + 3) + 1 (2x + 3)
= (2x + 3) (x + 1)
x² -1 = (x + 1) (x – 1)

Question 10.
Simplify

Answer:
(x² – x – 6) = (x – 3) (x + 2)
2x² + 5x – 3 = 2×2 + 6x – x – 3

= 2x (x + 3) -1 (x + 3)
= (x + 3) (2x – 1)
x² + 5x + 6 = (x + 2) (x + 3)

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)

6x² – x – 2 = 6x² – 4x + 3x – 2
= 2x (3x – 2) + 1 (3x – 2)
= (3x – 2) (2x + 1)

4x² + 8x + 3 = 4x² + 6x + 2x + 3
= 2x(2x + 3) + 1 (2x + 3)
= (2x + 3) (2x + 1)

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:

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,

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”

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:

6x² – 12x + 9x – 18 = 0
6x(x – 2) + 9(x – 2) = 0

(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

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

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

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)

Since the width cannot be negative. Width of the garden = 2 m

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



(iii) \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\alpha \beta}\)

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

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

Question 20.
Find X and Y if

Answer:

Question 21.
Solve for x,y

Answer:

Question 22.

Show that A² – 7A + 101₃ = 0
Answer:

Question 23.
Verify that (AB)^T = B^T A^T if

Answer:

From (1) and (2) we get
(AB)^T = B^TA^T

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.

(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


(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}.

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.

(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

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}.