Algebra

Question: For what values of m are the roots of the quadratic equation mx (x - 2√5) + 10 = 0 real and equal?

Solution:
mx (x - 2√5) + 10 = 0
⇒ mx² - 2√5 mx + 10=0

Compare given equation with the general form of quadratic equation, which ax² + bx + c=0
a = m, b = - 2√5m, c = 10

Since roots are real and equal, discriminant, D = 0
b² - 4ac = 0
⇒ (- 2√5m)² - 4 × m × 10 = 0
⇒ 20m² - 40m = 0
⇒ 20(m² - 2m) = 0
⇒ m² - 2m = 0
⇒ m(m - 2) = 0

Either, m = 0 

Or, m - 2 = 0
∴ m = 2

Question:

Solution:

Question: In a class, 30 students play basketball, 20 students play volleyball, and 8 students play both. 12 students play neither basketball nor volleyball. What is the total number of students in the class?

Solution:
Let the number of students who play basketball = 30
Number of students who play volleyball = 20
Number of students who play both basketball and volleyball = 8
Number of students who play neither = 12

First, calculate the number of students who play basketball or volleyball:
n(B ∪ V) = n(B) + n(V) − n(B ∩ V)
n(B ∪ V) = 30 + 20 − 8 = 42

Now, add the students who play neither sport to get total students:
Total students = n(B ∪ V) + neither

Total students = 42 + 12 = 54

Question: If log₁₀4 + log₁₀(3x + 30)= log₁₀(2x + 8) + 1, then what is the value of x?

Solution:
Given equation:
⇒ log₁₀4 + log₁₀(3x + 30) = log₁₀(2x + 8) + 1
⇒ log₁₀4 + log₁₀(3x + 30) = log₁₀(2x + 8) + log₁₀10  [যেহেতু, log₁₀10 = 1]
⇒ log₁₀[4(3x + 30)] = log₁₀[10(2x + 8)]  [যেহেতু, logA + logB = log(AB)]
⇒ 4(3x + 30) = 10(2x + 8)
⇒ 12x + 120 = 20x + 80
⇒ 20x − 12x = 120 − 80
⇒ 8x = 40
⇒ x = 5

Let, Side of square = x
Here, √2x = 20
Or, 2x² = 400
⇒ x² = 200
⇒ x = 14.14
So, perimeter = 4 × 14.14 = 56.56 ≅ 60 [As the approximate value was asked]

Question: If (4P + 1)² = 441, then P³/3P = ?

Solution:
(4P + 1)² = 441
or, (4P + 1) = √441
or, 4P + 1 = 21
or, 4P = 21 - 1
or, 4P = 20
or, P = 20/4
∴ P = 5

∴ P³/3P = 5³/(3 × 5)
= 125/15
= 25/3

Question: If |2x - 3| ≤ 9, then which of the following intervals represents all possible values of the expression 5x + 7?

Solution:
Start with the given inequality:
|2x - 3| ≤ 9

Rewrite as a compound inequality:
- 9 ≤ 2x - 3 ≤ 9
⇒ - 9 + 3 ≤ 2x - 3 + 3 ≤ 9 + 3
⇒ - 6 ≤ 2x ≤ 12
⇒ - 3 ≤ x ≤ 6

Now, find the range of 5x + 7:
Multiply the interval by 5:
5(- 3) ≤ 5x ≤ 5(6) 
⇒ - 15 ≤ 5x ≤ 30
⇒ - 15 + 7 ≤ 5x + 7 ≤ 30 + 7 
⇒ - 8 ≤ 5x + 7 ≤ 37

So, all possible values of 5x + 7 lie in the interval: [ - 8, 37 ]

Let, The numbers are x & y,
therefore, x - y = 11 ---- (1) and
1/5 (x + y) = 9
or, x + y = 45 ------ (2)
Adding two equation we got,
2x = 56 or, x = 28
Putting the value of x in equation 1,
We get, y = 17

Question: What is the 9^th term of the sequence : - 2, - 4, - 6, ............................ , - 100?

Solution:
Here,
- 4 - (- 2) = - 4 + 2 = - 2
- 6 - (- 4) = - 6 + 4 = - 2
∴ d = - 2
a = - 2
n = 9

∴ The 9^th term of the sequence = a + (n - 1)d
= - 2 + (9 - 1) (- 2)
= - 2 + 8 (- 2)
= - 2 - 16
= - 18

√(10 + √(25 + √(108 + √(154 + √225))))
= √(10 + √(25 + √(108 + √(154 + 15))))
= √(10 + √(25 + √(108 + √169)))
= √(10 + √(25 + √(108 + 13)))
= √(10 + √(25 + √121))
= √(10 + √(25 + 11))
= √(10 + √36)
= √(10 + 6)
= √16
= 4

Given, x + 1/x = 3
x³  + 1/x³  = (x + 1/x)³ - 3.x.1/x(x + 1/x)
= (3)³ - 3(3)
= 18

Question: The value of 
  is = ?

Solution: 

Question: Solve the inequality |x - 2| < 5

Solution:
|x - 2| < 5
⇒ - 5 < x - 2 < 5
⇒ - 5 + 2 < x - 2 + 2 < 5 + 2
⇒ - 3 < x < 7

Question: If 5x - (5/x) = 15, then what is the value of x³ - (1/x)³ ?

Solution:
দেওয়া আছে,
5x - 5/x = 15
⇒ (5x - 5/x) / 5 = 15 / 5
∴ x - 1/x = 3

এখন,
x³ - (1/x)³
= (x - 1/x)³ + 3 × x × (1/x)(x - 1/x)
= (x - 1/x)³ + 3(x - 1/x)
= 3³ + 3 × 3
= 27 + 9
= 36

x = 1 + √2 + √3 ...........(i)
y = 1 + √2 - √3 .............(ii)

x² + 4xy + y²/(x + y)
= {(x + y)² + 2xy}/(x + y)

From (i) + (ii)
x + y = 2 + 2√2
xy = (1 + √2)² - (√3)²
= 3 + 2√2 - 3
= 2√2

{(x + y)² + 2xy}/(x + y)
= {(2 + 2√2)² + (2 × 2√2)}/(2 + 2√2)
= (12 + 12√2)/(2 + 2√2)
= 12(1 + √2)/2((1 + √2)
= 12/2
= 6.

Question: If x = 2 + √5 and y = 2 - √5, find the value of x² + y².

Solution:
দেওয়া আছে,
x = 2 + √5
y = 2 - √5

∴ x + y = (2 + √5) + (2 - √5) = 4

এবং
xy = (2 + √5)(2 - √5)
= 2² - (√5)²
= 4 - 5
= - 1

এখন,
x² + y²
= (x + y)² - 2xy
= (4)² - 2(- 1)
= 16 + 2
= 18

Question: The distance between two points (- 6, y) and (18, 6) is 26 units. Find the value of y.

Solution:
Given that, 
The distance between two point = 26 units
The value of the first co-ordinate = (x₁, y₁) = (- 6, y)
The value of the second co-ordinate = (x₂, y₂) = (18, 6)

We know, 
Distance = √{(x_2​ - x₁​)² + (y₂​ - y_1​)²}

According to the question,
26 = √{(18 - (- 6))² + (6 - y)²
⇒ 24² + (6 - y)² = 26²  ;[Squaring on both sides of the equation.]
⇒ 576 + (6 - y)² = 676
⇒ (6 - y)² = 100 = 10²
⇒ 6 - y = 10
⇒ y = 6 - 10
∴ y = - 4

∴ The required answer is - 4.

Here, Second term is = a + (2 - 1)d = 4
∴ a + d = 4 ...... (i)
and, Tenth term is = a + (10 - 1) = 15
∴ a + 9d = 15 ...... (ii)
Now, (i)×9 - (ii),
8a = 21
∴ a = 2.625 ≅ 2.63

2x² + 3x - 2 = 0
Or, 2x² + 4x - x - 2 = 0
Or, 2x(x+2) - 1(x+2) = 0
Or, (2x - 1)(x + 2) = 0
Either, (2x - 1) = 0 Or, (x + 2) = 0
x = 1/2, -2

Question: If x⁴ ≤ 16 and y² ≤ 36, then the maximum possible value of (x - y) is:

Solution: 
Given that, 
x⁴ ≤ 16 
⇒ x⁴ ≤ 2⁴
⇒ x ≤ 2
∴ -2 ≤ x ≤ 2
And
y² ≤ 36
⇒ y² ≤ 6²
⇒ y ≤ 6
∴ - 6 ≤ y ≤ 6

Now, 
x could be positive 2 or negative 2. y could be positive 6 or negative 6 . The four possible values for (x - y) are as follows, 
1. 2 - 6 = - 4  ; [When x = 2 and y = 6]
2. 2 - (- 6) = 2 + 6 = 8  ; [When x = 2 and y = - 6]
3. - 2 - 6 = - 8  ; [When x = - 2 and y = 6]
4. - 2 - (- 6) ​= - 2 + 6 = 4  ; [When x = - 2 and y = - 6]

So the maximum value would be 8.

Question: If x = 1 + √2 and y = 1 - √2, find the value of (x² + y²)².

Solution: 
Given that, x = 1 + √2 and y = 1 - √2
∴ x + y = 1 + √2 + 1 - √2
= 2

And, xy = (1 + √2)(1 - √2)
= 1² - (√2)²
= 1 - 2
= - 1

Now,
x² + y² = (x + y)² - 2xy
= (2)² - 2(- 1)
= 4 + 2
= 6

∴ (x² + y²)² = 6²
= 36

Let, Tanks capacity is x gallons
ATQ,
5x/8 - x/4 = 6
⇒ 3x/8 = 6
∴ x = (6×8)/3 = 16 gallons

Question: If 5 ≥ p ≥ - 1 and q ≥ - 1, which of the following cannot be a value of p - q?

Solution: 
Here, 5 ≥ p ≥ - 1 and q ≥ - 1

Now,
i) If, p = - 1 and q = - 1 then, p - q = - 1 - (- 1) = -1 + 1 = 0 
ii) If, p = 2 and q = 1 then, p - q = 2 - 1 = 1 
iii) If, p = 5 and q = 0 then, p - q = 5 - 0 = 5 
iv) If, p = 5 and q = -1 then, p - q = 5 - (- 1) = 5 + 1 = 6

All the given values are possible because the maximum value of p - q is 6, and 0, 1, 5, and 6 are all ≤ 6.

Correct answer: ঙ) None of these

Question: If x is equal to 2 more than the product of 4 and z, and y is equal to 3 less than the product of 5 and z, then 3x is how much greater than 2y when z is 3?

Solution: 
Given:
x = 4z + 2
y = 5z - 3

When z = 3,
∴ x = 4 × 3 + 2 = 12 + 2 = 14
∴ y = 5 × 3 - 3 = 15 - 3 = 12

Now,
∴ 3x = 3 × 14 = 42
∴ 2y = 2 × 12 = 24

∴ Difference = 3x - 2y = 42 - 24 = 18
∴ 3x is 18 greater than 2y.

দেয়া আছে,
k = 462/n যেখানে k একটি পূর্ণসংখ্যা।
∴ n এর মান এমন হবে যা দ্বারা 462 কে নিঃশেষে ভাগ করা যাবে। অপশন অনুযায়ী একমাত্র 22 দ্বারা 462 কে ভাগ করা যায়।

সুতরাং n এর মান 22.

Question: If (x - 2) is a factor of the polynomial x³ + 4x² - px + 10, what is the value of p?

Solution:
Let f(x) = x³ + 4x² - px + 10
Since (x - 2) is a factor of f(x),
by the Factor Theorem,
When x - 2 = 0 ⇒ x = 2, then f(x) = 0

Now,
f(2) = (2)³ + 4(2)² - p(2) + 10
= 8 + 4(4) - 2p + 10
= 8 + 16 - 2p + 10
= 34 - 2p

According to the condition,
f(2) = 0
⇒ 34 - 2p = 0
⇒ 2p = 34
⇒ p = 34/2
⇒ p = 17

So the value of p is 17.

3x + 2y = 7 .... (i)
3x - 2y = 5 .... (ii)

(i) + (ii), 6x = 12
Or, x = 2

From, (i), y = 1/2

So, xy = 2.1/2 = 1

[মূল প্রশ্নে Px = Qy = Rz এর পরিবর্তে P^x = Q^y = R^z হবে]

Question: If Px = Qy = Rz and Q/P = R/Q then 2z/(x + z) = ?

Solution:
ধরি,
P^x = Q^y = R^z = k
এখন,
P^x = k
∴ P = k^(1/x)

অনুরুপভাবে,
Q^y = k
∴ Q = k^(1/y)
এবং
R^z = k
∴ R = k^(1/z)

আবার,
⇒ Q/p = R/Q
⇒ Q² = PR
⇒ {k^(1/y)}² = k^(1/x) × k^(1/z)
⇒ k^(2/y) = k^{(z + x)/xz}
⇒ 2/y = (z + x)/xz
⇒ 2xz = y(z + x)
∴ 2z/x + z = y/x