ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: If x² - 45x + 324 = 0, then what is the values of x?

Solution:
x² - 45x + 324 = 0
⇒ x²- 36x - 9x + 324 = 0
⇒ x(x - 36) - 9(x - 36) = 0
⇒ (x - 36)(x - 9) = 0
Either x - 36 = 0 or x - 9 = 0
∴ x = 36, 9

Question: If a² + b² + c² = 16 and ab + bc + ca = 10, find the value of a + b + c.

Solution:
a² + b² + c² = 16
ab + bc + ca = 10

We know that,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (a + b + c)² = 16 + 2 × 10
⇒ (a + b + c)² = 36
⇒ a + b + c = √36
⇒ a + b + c = ± 6

∴ The value of (a + b + c) is ± 6.

Question: What must be added to the polynomial f(x) = x⁴ + 2x³ - 2x² + x - 1 so that the resulting polynomial is exactly divisible by x² + 2x - 3?

Solution:
x² + 2x - 3 ) x⁴ + 2x³ - 2x² + x - 1( x² + 1
                    x⁴ + 2x³ - 3x² 
                  ________________________
                                     x² + x - 1
                                     x² + 2x - 3
                  ________________________
                                          - x + 2

To get exactly divisible, the remainder must be 0
- x + 2 + k = 0
⇒ k = (x - 2)

Hence, the correct option is 1.

Question: If x - 1/x = 3, the value of x³ - 1/x³ is-

Solution:
x - 1/x = 3

x³ - 1/x³
= (x - 1/x)³ + 3.x.(1/x)(x - 1/x)
= (x - 1/x)³ + 3(x - 1/x) 
= (3)³ + 3 × 3
= 27 + 9
= 36

Question: Find the degree of the polynomial 2x⁵ + 2x³y³ + 4y⁴ + 5.

Solution:
The degree of a polynomial is the highest of the degrees of its individual terms with non-zero coefficients.

Degree of the polynomial in 2x⁵ = 5
Degree of the polynomial in 2x³y³ = 6
Degree of the polynomial in 4y⁴ = 4
Degree of the polynomial in 5 = 0

Hence, the highest degree is 6
∴ Degree of polynomial = 6

Question: The sum of values of x satisfying x^2/3 + x^1/3 = 2 is-

Solution:
x^2/3 + x^1/3 = 2
⇒ (x^2/3 + x^1/3)³ = 2³
⇒ (x^2/3)³ + (x^1/3)³ + 3.x^2/3.x^1/3(x^2/3 + x^1/3) = 8
⇒ x² + x + 3x(x^2/3 + x^1/3) = 8
⇒ x² + x + 3x(2) = 8
⇒ x² + 7x - 8 = 0
⇒ x² + 8x - x - 8 = 0
⇒ x (x + 8) - 1 (x + 8) = 0
⇒ (x + 8)(x - 1) = 0 
⇒ x = - 8 or x = 1

∴ Sum of values of x = - 8 + 1 = - 7.

Question: Quadratic equation corresponding to the roots 2 + √5 and 2 - √5 is-

Solution:
The quadratic equation is: x² - (Sum of roots)x + Product of roots = 0

Let the roots of the equation be A and B.
A = 2 + √5 and B = 2 - √5

∴ A + B = 2 + √5 + 2 - √5 = 4

∴ A × B = (2 + √5)(2 - √5) = 4 - 5 = - 1

Then equation is
x² - 4x - 1 = 0

Question: Find the product of two consecutive numbers where four times the first number is 10 more than thrice the second number.

Solution:
Suppose the numbers are 'a' and 'a + 1'.
According to the question :
4a = 3(a + 1) +10
⇒ 4a = 3a + 3 + 10
∴ a = 13

Hence, the numbers are 13 and 14.
∴ Product = 13 × 14 = 182

Question: If a + b + c = 2s, then [(s - a)² + (s - b)² + (s - c)² + s²] =?

Solution:
[(s - a)² + (s - b)² + (s - c)² + s²]
= (s² + a² - 2as) + (s² + b² - 2sb) + (s² + c² - 2sc) + s²
= 4s² + (a² + b² + c²) - 2s(a + b + c)
= 4s² + a² + b² + c² - 4s²
= a² + b² + c²

Question: If x = √10 + 3 then find the value of x³ - 1/x³.

Solution:
x = √10 + 3
1/x = 1/(√10 + 3)
=(√10 - 3) /{(√10 + 3)(√10 - 3)}
= (√10 - 3)/{(√10)² - (3)²}
= (√10 - 3)/(10 - 9)
= (√10 - 3)

x - 1/x = √10 + 3 -  (√10 - 3)
= √10 + 3 - √10 + 3
= 6

x³ - 1/x³
= (x - 1/x)³ + 3.x.(1/x)(x - 1/x)
= (x - 1/x)³ + 3(x - 1/x) 
= (6)³ + 3 × 6
= 216 + 18
= 234

Question: Solve the equation x² + 4x - 5 = 0.

Solution:
x² + 4x - 5 = 0
⇒ x² - 1x + 5x - 5 = 0
⇒ x(x - 1) + 5(x - 1) = 0
⇒ (x - 1)(x + 5) = 0
Hence, (x - 1) = 0, and (x + 5) =0
⇒ x - 1 = 0
∴ x = 1

similarly, x + 5 = 0
∴ x = - 5.

Therefore,
x = - 5 & x = 1

Question: Nine times a whole number is equal to five less than twice the square of the number. Find the number?

Solution:
Let the required whole number be x.

According to the question,
9x = 2x² - 5
⇒ 2x² - 9x - 5 = 0
⇒(x - 5)(2x + 1) = 0
⇒ x - 5 = 0 or 2x + 1 = 0
⇒ x = 5 or x = - 1/2

Since x is supposed to be a whole number, the answer, i.e., the required whole number is 5.

Question: Determine the roots of the equation (x + 3)(x - 3) = 40.

Solution:
Given,
(x + 3) (x - 3)=40
⇒ x² - 9 = 40
⇒ x² - 9 - 40 = 0
⇒ x² - 49 = 0
⇒ x² - 7² = 0
⇒ (x+7) (x-7) = 0
⇒ x + 7 = 0 or x - 7 = 0
∴ x = - 7 or x = 7.
Hence, the roots of the given equation are -7, 7.

Question: Solve the following quadratic equation by factoring.
z² - 16z + 61 = 2z - 20

Solution:
z² - 16z + 61 = 2z - 20
⇒ z² - 18z + 81 = 0
⇒ (z - 9)² = 0
∴ z = 9

Question: If p(x) = 3x⁴ - 2x² + x - 1, q(x) = 7x⁵ + 2x², then find the value of p(x) + q(x).

Solution:
p(x) + q(x)
= 3x⁴ - 2x² + x - 1 + 7x⁵ + 2x²
= 7x⁵ + 3x⁴ + x - 1

Question: Find the remainder when p(x) = x⁴ - 3x² - 10x + 2 is divided by (x - 3).

Solution:
The remainder is p(3)

p(x) = x⁴ - 3x² - 10x + 2
∴ p(3) = 3⁴ - 3.3² - 10.3 + 2
= 81 - 27 - 30 + 2
= 26

Question: What are the roots of the equation √(2x + 9) = 13 - x?

Solution:
√(2x + 9) = 13 - x
Squaring both sides, we get
2x + 9 = (13 - x)²
⇒ 2x + 9 = 169 - 26x + x²
⇒ - x² + 28x - 160 = 0
⇒ x² - 28x + 160 = 0
⇒ (x - 8)(x - 20) = 0
⇒ x - 8 = 0 or x - 20 = 0
⇒ x = 8 or x = 20

But x = 20 does not satisfy the given equation, so it is rejected. Hence, the root of the given equation is 8.

Question: Rafiul has more marbles than Roman and they have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with.

Solution:
Let
The number of marbles Rafiul had be x.
Then the number of marbles Roman had = 45 - x.

The number of marbles left with Rafiul after losing 5 marbles = x - 5
The number of marbles left with Roman after losing 5 marbles = 45 - x - 5 = 40 - x

ATQ,
(x - 5) (40 - x) = 124
⇒ 40x - x² - 200 + 5x = 124
⇒ - x² + 45x - 200 = 124
⇒ x² - 45x + 324 = 0
⇒ x² - 36x - 9x + 324 = 0
⇒ x(x - 36) - 9(x - 36) = 0
⇒ (x - 36)(x - 9) = 0
∴ x = 36 and x = 9

So, the number of marbles Rahul had is 36 and Rohan had is 9.

Question: If a + b + c = 0, the value of a²/(bc) + b²/(ca) + c²/(ab) is-

Solution: 
a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

a + b + c = 0,
then a³ + b³ + c³ - 3abc = 0
∴ a³ + b³ + c³ = 3abc

a²/(bc) + b²/(ca) + c²/(ab)
= (a³ + b³ + c³)/abc
= (3abc)/(abc)
= 3 

Question: The polynomial p(x) = x⁵ - 7x³ + ax + 1 has remainder 13 after division by x - 1. Find the value of the coefficient a.

Solution:
p(x) = x⁵ - 7x³ + ax + 1
p(1) = 1⁵ - 7.1³ + a.1 + 1
= 1 - 7 + a + 1
= a - 5

∴ a - 5 = 13
∴ a = 18