৪৫তম বিসিএস, প্রাইমারি শিক্ষক নিয়োগ, সরকারি চাকরি, ব্যাংক প্রস্তুতি

Algebra (71 টি প্রশ্ন )

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Q1. (x × x² × x³ × x⁴ × x⁵) ÷ x⁸ = ?

ক) x⁶

খ) x⁷

গ) x³

ঘ) x⁹

ব্যাখ্যা (Explanation):

(x × x² × x³ × x⁴ × x⁵) ÷ x⁸₌ x^1+2+3+4+5÷x⁸=x^15-8
=x⁷

Q2. (1000)^y/3 = 10 হলে, y এর মান কত?

ক) 2

খ) 1

গ) 3

ঘ) 1.5

ব্যাখ্যা (Explanation):

(1000)^y/3=10

(10³)^y/3=10

10^y=10¹

Y=1

Q3. a²b³/c²d কে a³b²/cd³ দ্বারা ভাগ করলে ভাগফল কত?

ক) ab²/ca

খ) ab²/cd

গ) b²c/ad

ঘ) bd²/ac

ব্যাখ্যা (Explanation):

a2b3/c2d / a3b2/cd3

⇒ a2b3/c2d × cd3/a3b2

⇒bd²/ac

Q4. (√7 + √7)²= কত-

ক) 98

খ) 49

গ) 28

ঘ) 21

ব্যাখ্যা (Explanation):

(2√7)² = (2√7) × (2√7)

= 4 × (√7 × √7)

= 4 × 7

= 28

So, (√7 + √7)² = 28.

Q5. (x - y)² = 2 এবং xy = 2 হলে x² + y² এর মান-

ক) 12

খ) 14

গ) 16

ঘ) 6

ব্যাখ্যা (Explanation):

x² + y² = ( x - y) ² + 2xy

= 2 + 2 × 2

= 2 + 4

= 6

Q6. If 2x= 3 and y/4= 3, then (3+y)/(x+4)= ?

ক) 30/11

খ) 30/9

গ) 6/11

ঘ) 3/10

ব্যাখ্যা (Explanation):

Given,
2x= 3
=> x= 3/2,

and, y/4= 3
=> y=12.

Now, (3+y)/(4+x)

= (3+12)/(4+(3/2))

=15/(8+3)/2)

=15× 2/11
=30/11

Q7. If a³ + 3a²b + 3ab² + b³ = 3√3, Which one of the following is the value of √(a + b)?

ক) ⁴√3

খ) 3

গ) 3 √3

ঘ) 3²

ব্যাখ্যা (Explanation):

a³ + 3a²b + 3ab² + b³ =3√3
or,(a + b)³ = 3 √3
or,a + b = ³√(3√3)
or,a + b =(3√3)^1/3
or,√(a+b)=√{(3√3)^1/3}
=(3√3) ^1/(3×2)
=(3√3)^1/6
={(√3)² × ×3}^1/6
={(√3)³}^1/6
={3^1/2}^1/2
=3^1/4
=⁴√3

Q8. A polynomial f(x) = x⁴ – 11x³ + 31x² – 46x + 20 is defined. When it is divided by x² – 3x + n, the remainder left is q. Find the product of n and q.

ক) 20

খ) 30

গ) 40

ঘ) 50

ঙ)

100

ব্যাখ্যা (Explanation):

Let the quotient after division be x² + ax + b.

⇒ (x² – 3x + n)( x² + ax + b) + q = x⁴ – 11x³ + 31x² – 46x + 20

⇒ x⁴ + (a – 3)x³ + (n – 3a + b)x² + (an – 3b)x + bn + q = x⁴ – 11x³ + 31x² – 46x + 20

Comparing, we get a = -8, n – 3a + b = 31, an – 3b = -46, bn + q = 20.

Put value of a, we get n + b = 7, 8n + 3b = 46

Solving, we get n = 5, b = 2.

Now, bn + q = 20

⇒ q = 20 – 5× 2 = 10

∴ Product of n and q will be 50.

Q9. On multiplying the polynomials x² + px + q and x² + mx + n with each other, we get a polynomial whose zeroes are 1, 2, 3 and 4. What will be the value of (p + m)(q + n)?

ক) -110

খ) -140

গ) -180

ঘ) -210

ঙ) Cannot be determined

ব্যাখ্যা (Explanation):

Here, p, m, q and n could attain different values in different cases. For example, if x² + px + q correspond to roots 1 and 2, and x² + mx + n correspond to roots 3 and 4, then p = -3, q = 2, m = -7, n = 12. Here, (p + m)(q + n) = (-10)(14) = -140. And, if x² + px + q correspond to roots 1 and 3, and x² + mx + n correspond to roots 2 and 4, then p = -4, q = 3, m = -6, n = 8. Here, (p + m)(q + n) = (-10)(11) = -110.

We see that unique value of (p + m)(q + n) cannot be determined.

এখানে প্রথম ৩০টি প্রশ্নের ব্যাখ্যা দেখতে পারবেন, বাকি সব প্রশ্নের সম্পূর্ণ ব্যাখ্যা পেতে এখনই অ্যাপ ইন্সটল করুন।

Install App

Q10. The age of my father is equal to sum of my mother’s age and my age. When I was born, the ratio of ages of my mother and father was 5 : 6. After 8 years, the ratio of my age and my mother’s age will be 3 : 8. What was my mother’s age when I was 2 years old?

ক) 22 years

খ) 26 years

গ) 24 years

ঘ) 32 years

ঙ) 30 years

ব্যাখ্যা (Explanation):

Let my age be M years and my mother’s age be N years.

⇒ Age of my father = (M + N) years

When I was born, the ratio of ages of my mother and father was 5 : 6.

I was born M years back.

⇒ (N - M)/(M + N - M) = 5/6

⇒ 6N - 6M = 5N

⇒ N = 6M

After 8 years, the ratio of my age and my mother’s age will be 3 : 8.

⇒ (M + 8)/(N + 8) = 3/8

⇒ 8M + 64 = 3N + 24

Put N = 6M

⇒ 8M + 40 = 18M

⇒ M = 40/10 = 4

⇒ N = 6M = 6 × 4 = 24

Mother’s age = 24 years

∴ I am 4 years old. When I was 2 years old, it would be 2 years back from now, my mother’s age was 22 years.

Q11. In the question, one or more equations(s) is (are) provided. On the basis of these you have to give answer (i) q² – 11q + 30 = 0 (ii) 2p² – 7p + 6 = 0

ক) p = q

খ) p > q

গ) q > p

ঘ) p ≥ q

ঙ) q ≥ p

ব্যাখ্যা (Explanation):

Equation 1:

q² – 11q + 30 = 0

⇒ q² – 6q – 5q + 30 = 0

⇒ q (q – 6) – 5(q – 6) = 0

⇒ (q – 5) (q – 6) = 0

⇒ q = 5 or q = 6

Equation 2:

2p² – 7p + 6 = 0

⇒ 2p² – 4p – 3p + 6 = 0

⇒ 2p(p – 2) – 3(p – 2) = 0

⇒ (2p – 3) (p – 2) = 0

⇒ p = 3/2 or p = 2

Thus, p values are less than q. So, p < q

Q12. Given below are two quantities named A and B. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose between the possible answers. Quantity A: If a number is as much greater than 53 as it is less than 175, then the number: Quantity B: If \(p = \frac{{5 - {q^2}}}{{p - q}}\) and the value of p³ + q³ = 45 then the value of (p + q)^3/2:

ক) Quantity A > Quantity B

খ) Quantity A < Quantity B

গ) Quantity A ≥ Quantity B

ঘ) Quantity A ≤ Quantity B

ঙ) Quantity A = Quantity B

ব্যাখ্যা (Explanation):

First we will find Quantity A,

Quantity A:

Let the number be ‘x’.

(x – 53) = (175 – x)

⇒ x + x = 175 + 53

⇒ x = 228/2 = 114

∴ x = 114

Now,

Quantity B:

\(p = \frac{{5 - {q^2}}}{{p - q}}\)

⇒ p (p – q) = 5 – q²

⇒ p² – pq + q² = 5

⇒ (p² – pq + q²) = 5

We know that,

a³ + b³ = (a + b)(a² – ab + b²)

Now,

p³ + q³ = (p + q)(p² – pq + q²)

⇒ 45 = (p + q) × 5

⇒ (p + q) = 45/5 = 9

∴ (p+q)^3/2 = 9^3/2 = 27

∴ Quantity A > Quantity B

Q13. Given below are two quantities named A and B. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose between the possible answers. 0 < x < 1 Quantity A: (3x³ - 9x) (5x + 7) Quantity B: (3x² + 9) (5x² + 7x)

ক) Quantity A > Quantity B

খ) Quantity A < Quantity B

গ) Quantity A ≥ Quantity B

ঘ) Quantity A ≤ Quantity B

ঙ) Quantity A = Quantity B OR relationship cannot be determined.

ব্যাখ্যা (Explanation):

Notice that in each of the quantities, we can factor 3x out of the given expressions.

0 < x < 1

Quantity A: (3x³ - 9x) (5x + 7) = 3x(x² - 3) (5x + 7)

Quantity B: (3x² + 9) (5x² + 7x) = 3x(x² + 3) (5x + 7)

Because we know that x is not 0, we can divide away 3x from both quantities.

Also, 0 < x < 1, so (5x + 7) ≠ 0, we can divide away (5x + 7) from both quantities as well.

Quantity A: 3x(x² - 3) (5x + 7) = x² - 3

Quantity B: 3x(x² + 3) (5x + 7) = x² + 3

Since x² will always be positive, (x² + 3) will always be bigger than (x² – 3).

Q14. Father’s present age is 50 yrs. His age was equal to his son’s present age at the time of the son’s birth. What was son’s age 10 yrs. ago?

ক) 12

খ) 15

গ) 20

ঘ) 25

ঙ) None of these

ব্যাখ্যা (Explanation):

Let the son’s present age be x years.

From the given data, it is known that

Father’s age at the time of son’s birth = Son’s present age

∴ 50 – x =x

⇒ 50 = 2x

⇒ x = 25

∴ son’s age 10 yrs ago = (25 - 10) = 15 yrs.

Q15. Direction: In the following questions two equations numbered I and II are given. You have to solve both the equations and find relation between x and y. I. 121x² – 17 = 8 II. \(13y + \sqrt 9 = \sqrt {64}\)

ক) if x > y

খ) if x ≥ y

গ) if x < y

ঘ) if x ≤ y

ঙ) if x = y or the relationship cannot be established.

ব্যাখ্যা (Explanation):

According to the given equations,

I. 121x² – 17 = 8

⇒ 121x² = 8 + 17 = 25

⇒ x = \(\sqrt {\frac{{25}}{{121}}}\)

∴ x = ± 5/11

II. \(13y + \sqrt 9 = \sqrt {64}\)

⇒ 13y + 3 = 8

⇒ 13y = 8 – 3 = 5

∴ y = 5/13

∵ The value of y is greater than -5/11 but less than 5/11

Hence, no relationship can be established

Q16. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘x’ and ‘y’ and mark correct answer. I. x² – 19x + 84 = 0 II. y² – 25y + 126 = 0

ক) If x > y

খ) If x ≥ y

গ) If x < y

ঘ) If x ≤ y

ঙ) If x = y or the relation cannot be determined

ব্যাখ্যা (Explanation):

We will solve both the equations separately.

Equation I:

x² – 19x + 84 = 0

⇒ x² – 12x – 7x + 84 = 0

⇒ (x – 12) (x – 7) = 0

⇒ x = 12 or x = 7

Equation II:

y² – 25y + 126 = 0

⇒ y² – 18y – 7y + 126 = 0

⇒ (y – 18) (y – 7) = 0

⇒ y = 18 or y = 7

So, the relation cannot be determined

Q17. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘p’ and ‘q’ and mark correct answer. 1. \(\frac{{135}}{{\sqrt p }} - 22\sqrt p = 23\sqrt p\) 2. \(\frac{{2 \times 324 \div \sqrt q }}{{108}} - 2\sqrt q = 0\)

ক) p > q

খ) p ≥ q

গ) p < q

ঘ) p ≤ q

ঙ) p = q or the relation cannot be determined

ব্যাখ্যা (Explanation):

We will solve both the equations separately.

Equation I:

\(\frac{{135}}{{\sqrt p }} - 22\sqrt p = 23\sqrt p\)

Multiplying both sides by √p, we get

⇒ 135 – 22p = 23p

⇒ 23p + 22p = 135

⇒ 45p = 135

⇒ p = 135/45

⇒ p = 3

Equation II:

\(\begin{array}{l} \frac{{2 \times 324 \div \sqrt q }}{{108}} - 2\sqrt q = 0\\ \Rightarrow \frac{{2 \times 324}}{{108 \times \sqrt q }} = 2\sqrt q \end{array}\)

Multiplying both sides by √q, we get

\(\Rightarrow q = \;\frac{{2 \times 324}}{{108 \times 2}}\)

⇒ q = 3

Comparing the values of p and q, we notice that p = q

Q18. Two hackers planned to hack a bank account which had debit card pin code in such a format that if a fraction could be made using the first two digits as numerator and last two digits as denominator, then if the numerator of the fraction was increased by 200% and the denominator by 300%, the fraction becomes 12/15. What was the pin of the debit card?

ক) 1889

খ) 1660

গ) 4850

ঘ) 1890

ঙ) 1615

ব্যাখ্যা (Explanation):

n percent (%) of a number x = \(x \times \frac{n}{{100}}\)

Suppose the original fraction to be x/y.

Then, \(\frac{{x+x \times \frac{{200}}{{100}}}}{{y+y \times \frac{{300}}{{100}}}} = \frac{{12}}{{15}}\)

\(\Rightarrow \frac{x}{y} = \frac{{12 \times 4}}{{15 \times 3}}\)

\(\Rightarrow \frac{x}{y} = \frac{{16}}{{15}}\)

∴ Debit card pin is 1615

Q19. Below question consists of two equations. On the basis of these two equations you have to find out the relation between p and q. I. \(p = \frac{{\sqrt 4 }}{{\sqrt 9 }}\) II. 9q² – 12q + 4 = 0

ক) p > q

খ) q > p

গ) p = q

ঘ) p ≥ q

ঙ) q ≥ p

ব্যাখ্যা (Explanation):

9q² – 12q + 4 = 0

⇒ (3q – 2)² = 0

q = 2/3

Also,\(p = \frac{{\sqrt 4 }}{{\sqrt 9 }}\)

Hence p = 2/3

Hence we have p = q

এখানে প্রথম ৩০টি প্রশ্নের ব্যাখ্যা দেখতে পারবেন, বাকি সব প্রশ্নের সম্পূর্ণ ব্যাখ্যা পেতে এখনই অ্যাপ ইন্সটল করুন।

Install App

Q20. The number of solutions of x² + 4|x| + 5 = 0 is

ক) 0

খ) 2

গ) 1

ঘ) 5

ঙ) 4

ব্যাখ্যা (Explanation):

|x| is always positive.

The given equation is x² + 4|x| + 5 = 0

∴ all the terms on the left hand side are positive.

Thus, it will never be equal to zero for any value of x.

∴ Number of roots of the given equation are 0.

Q21. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘l’ and ‘m’ and mark correct answer. I. (125)^1/3 l – (1331)^1/3 = 4 × (729)^1/6 + (1024)^1/10 II. (m)² – (15625)^1/3 = 0

ক) l > m

খ) l ≥ m

গ) l < m

ঘ)

l ≤ m

ঙ) l = m or the relation cannot be determined

ব্যাখ্যা (Explanation):

We solve both equations separately

Equation I:

(125)^1/3 l – (1331)^1/3 = 4 × (729)^1/6 + (1024)^1/10

⇒ 5l – 11 = 4 × 3 + 2

⇒ 5l = 14 + 11

⇒ l = 25/5

⇒ l = 5

Equation II:

(m)² – (15625)^1/3 = 0

⇒ (m)² = (15625)^1/3

⇒ m² = 25

⇒ m² – 25 = 0

By using the formula: a² – b² = (a + b) (a – b)

⇒ m = ± 5

Comparing values of l and m, we obtain

l ≥ m

Q22. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘p’ and ‘q’ and mark correct answer. I. \({\left( {\frac{{1024 \times 36}}{{900}}} \right)^p} = \frac{{32}}{5}\) II. q²+ 2 × √2 × √2 × q + 2 × √2 × √2 = 0

ক) p > q

খ) p ≥ q

গ) p < q

ঘ) p ≤ q

ঙ) p = q or the relation cannot be determined

ব্যাখ্যা (Explanation):

We will solve both the equations separately.

Equation I:

\(\begin{array}{l} {\left( {\frac{{1024 \times 36}}{{900}}} \right)^p} = \frac{{32}}{5}\\ \Rightarrow \;{\left( {\frac{{{2^{10}} \times 1}}{{25}}} \right)^p} = \frac{{{2^{10\; \times \frac{1}{2}}}}}{{{{25}^{\frac{1}{2}}}}}\\ \Rightarrow \;\frac{{{2^{10 \times p}}}}{{{{25}^p}}} = \frac{{{2^{10\; \times \frac{1}{2}}}}}{{{{25}^{\frac{1}{2}}}}} \end{array}\)

On comparing both sides, we get

⇒ p = 1/2

Equation II:

q² + 2 × √2 × √2 × q + 2 × √2 × √2 = 0

⇒ q² + 2 × 2× q + 22 = 0

By using formula: (a + b)²= a² + 2ab + b²

⇒ (q + 2)2 = 0

⇒ q = -2, -2

Comparing the values of p and q we get,

p > q

Q23. A boy was asked to find 8/9th of a fraction. He made a mistake of dividing the fraction by 8/9 and so got an answer which exceeds the correct answer by 17/54. Find the original fraction?

ক) 2/3

খ) 3/4

গ) 4/3

ঘ) 3/5

ঙ) None of these

ব্যাখ্যা (Explanation):

Let the fraction be a/b

We consider fraction a/b = x

A boy was asked to find 8/9 th of a fraction. He made a mistake of dividing the fraction by 8/9 and so got an answer which exceeds the correct answer by 17/54

\(\Rightarrow \frac{x}{{\frac{8}{9}}}\;-\frac{8}{9} \times x\; = \frac{{17}}{{54}}\)

⇒ 9x/8 – 8x/9 = 17/54

\(\begin{array}{l} \Rightarrow \;\frac{{9x \times 9 - 8x \times 8}}{{72}} = \;\frac{{17}}{{54}}\\ \Rightarrow {\rm{\;}}\frac{{81x - 64x}}{{72}} = \frac{{17}}{{54}} \end{array}\)

⇒ 17x/72 = 17/54

\(\Rightarrow {\rm{\;x\;}} = \frac{{17}}{{54}}\; \times \frac{{72}}{{17}}\)

⇒ x = 4/3

So, our fraction a/b = x = 4/3

Q24. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘x’ and ‘y’ and mark correct answer. I. 9x – 15.45 = 54.55 + 4x II. \(\sqrt {y + 155} - \sqrt {36} = \sqrt {49}\)

ক) x > y

খ) x ≥ y

গ) x < y

ঘ) x ≤ y

ঙ) x = y or the relation cannot be determined

ব্যাখ্যা (Explanation):

We will solve both the equations separately.

Equation I:

9x – 15.45 = 54.55 + 4x

⇒ 9x – 4x = 70

⇒ 5x = 70

⇒ x = 14

Equation II:

\(\begin{array}{l} \sqrt {y + 155} - \sqrt {36} = \sqrt {49} \\ \Rightarrow \sqrt {y + 155} - 6 = 7\\ \Rightarrow \sqrt {y + 155} = 13 \end{array}\)

On squaring both sides,

y + 155 = 169

⇒ y = 169 – 155

⇒ y = 14

Comparing the values of x and y, we get,

x = y

Q25. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘x’ and ‘y’ and mark correct answer. I. \(\frac{{15 \times 4}}{{{x^{4/7}}}} - \frac{{8 \times 3}}{{{x^{4/7}}}} = {x^{10/7}}\) II. y³ + 783 = 999

ক) x > y

খ) x ≥ y

গ) x < y

ঘ) x ≤ y

ঙ) x = y or the relation cannot be determined

ব্যাখ্যা (Explanation):

We will solve both the equations separately.

Equation I:

\(\begin{array}{l} \frac{{15 \times 4}}{{{x^{\frac{4}{7}}}}} - \frac{{8 \times 3}}{{{x^{\frac{4}{7}}}}} = {x^{\frac{{10}}{7}}}\\ \Rightarrow \;\frac{{60}}{{{x^{4/7}}}} - \frac{{24}}{{{x^{4/7}}}} = {x^{10/7}}\\ \Rightarrow \;\frac{{36}}{{{x^{4/7}}}} = {x^{10/7}}\\ \Rightarrow \;36 = {x^{\frac{{10}}{7} + \frac{4}{7}}}\\ \Rightarrow \;36 = {x^2} \end{array}\)

⇒ x = ± 6

Equation II:

y³ + 783 = 999

⇒ y³ = 999 – 783

⇒ y³ = 216

⇒ y = 6

Comparing the values of x and y, we get,

x ≤ y

Q26. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘a’ and ‘b’ and mark correct answer. I. a√25 + 2b = √961 II. 3a + (2401)^1/4b = 36

ক) a > b

খ) a ≥ b

গ) a < b

ঘ) a ≤ b

ঙ) a = b or the relation cannot be determined

ব্যাখ্যা (Explanation):

We will solve both the equations separately.

Equation I:

a√25 + 2b = √961

Here √25 can only be evaluated as 5

⇒ 5a + 2b = 31

Equation II:

3a + (2401)^1/4b = 36

⇒ 3a + 7b = 36

Multiplying the simplified values of equation (I) by 3 and that of (II) by 5, we get,

b = 3

Putting this value of y in the simplified value of equation (I), we get

a = 5

∴ a > b

Q27. In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘l’ and ‘m’ and mark correct answer. I. (729)^1/3 l ÷ 423 = 1 II. 6 × (3375)^1/3m – (3240000)^1/2 = 36

ক) l > m

খ) l ≥ m

গ) l < m

ঘ) l ≤ m

ঙ) l = m or the relation cannot be determined

ব্যাখ্যা (Explanation):

We will solve both the equations separately.

Equation I:

(729)^1/3 l ÷ 423 = 1

\(\Rightarrow {\rm{\;}}\frac{{{{729}^{1/3}}}}{{423}}l\; = \;1\)

⇒ 9l = 423

⇒ l = 423/9 = 47

Equation II:

6 × (3375)^1/3m – (3240000)^1/2 = 36

⇒ 6 × 15m – 1800 = 36

⇒ 90m = 36 + 1800

⇒ m = 1836/90 = 20.4

Comparing the values of l and m we get,

l > m

Q28. In the following question, one or two equation(s) is/are given. On their basis, you have to determine the relation between p and q and then give answer I. 2p² = 23p – 63 II. 2q (q^-8) = q^-36

ক) p = q

খ) p > q

গ) p < q

ঘ) p ≤ q

ঙ) p ≥ q

ব্যাখ্যা (Explanation):

Equation 1:

2p² = 23p – 63

⇒ 2p2 - 14p – 9p + 63 = 0

⇒ 2p (p- 7) -9 (p - 7) = 0

⇒ (2p - 9) (p – 7) = 0

⇒ 2p – 9 = 0 and p – 7 = 0

⇒ p = 9/2 and 7

Equation 2:

2q (q^-8) = q^-36

⇒ 2q^-7 = q^-36

⇒ 2 q^-7 × q³⁶ = 1

⇒ q²⁹ = ½

⇒ q = (1/2)²⁹

Clearly, q < 1

And hence: p > q

Q29. In the following question, one or two equation(s) is/are given. On their basis you have to determine the relation between x and y and then give answer I. 3x² + 7x = 6 II. 6(2y² + 1) = 17y

ক) x < y

খ) x > y

গ) x ≤ y

ঘ) x ≥ y

ঙ) x = y

ব্যাখ্যা (Explanation):

3x² + 7x = 6

⇒ 3x² + 7x - 6 = 0

⇒ 3x² + 9x - 2x - 6 = 0

⇒ 3x (x + 3) - 2 (x + 3) = 0

⇒ (3x - 2) (x + 3) = 0

⇒ 3x - 2 = 0 or x + 3 = 0

⇒ x= 2/3 or -3

Equation 2:

6(2y² + 1) = 17y

⇒ 12y² + 6 = 17y

⇒ 12y² – 17y + 6 = 0

⇒12y² – 9y – 8y + 6 = 0

⇒ 3y (4y- 3) – 2 (4y - 3) = 0

⇒ (3y- 2) (4y - 3) = 0

⇒ 3y – 2 = 0 and 4y - 3 = 0

⇒ y = 2/3 and 3/4

Comparing values of x and y

x ≤ y

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Q30. In the following question,two equations are given. On their basis you have to determine the relation between x and y and then give answer I. 3x² – 7x + 2 = 0 II. 2y² – 11y + 15 = 0

ক) x < y

খ) x > y

গ) x ≤ y

ঘ) x ≥ y

ঙ) x = y

ব্যাখ্যা (Explanation):

3x² - 7x + 2 = 0

⇒ 3x²- 6x - x + 2 = 0
⇒ 3x (x - 2) - 1 (x - 2) = 0

⇒ (3x - 1) (x - 2) = 0

⇒ 3x - 1 = 0
⇒ 3x = 1
⇒ x = 1/3
or,
x - 2 = 0
⇒ x = 2

Equation 2:

2y² - 11y + 15 = 0
⇒ 2y² - 6y - 5y +15 = 0
⇒ 2y (y - 3) - 5 (y - 3) = 0
⇒ (2y - 5) (y - 3) = 0
⇒ (2y - 5) = 0
⇒ y = 5/2
Also (y - 3) = 0
⇒ y = 3

Now comparing x and y:

x = 1/3 is smaller than 5/2 and 3

x= 2 is smaller than 3 and 5/2

Thus we see that x is smaller than y

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Algebra (71 টি প্রশ্ন )