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Concept

The difference between the consecutive square number is the sum of the two numbers that are squared.

Calculation

(10² - 9²) + (12² - 11²) + (14² - 13²)

⇒ 19 + 23 + 27

⇒  69

Given:

\(a-\dfrac{1}{a}=7\)

Formula used:

(a – b)³ = a³ – b³ – 3ab(a – b)

Calculation:

\(a-\dfrac{1}{a}=7\)      ----1

⇒  \({\left( {a - \;\frac{1}{a}} \right)^3} = {a^3} - \;\frac{1}{{{a^{3\;}}}}\; - \;\frac{{3a}}{a}\left( {a - \;\frac{1}{a}} \right)\)

⇒  \({a^3} - \;\frac{1}{{{a^{3\;}}}}\; = \;{\left( {a - \;\frac{1}{a}} \right)^3} + 3\left( {a - \;\frac{1}{a}} \right)\;\)

⇒ a³ – 1/a³ = 7³ + 3 × 7

⇒ 343 + 21

⇒ 364

∴ \({a^3} - \;\frac{1}{{{a^{3\;}}}}\; = 364\;\)

Given:

\(\dfrac{5x-7}{3}+2=\dfrac{4x-3}{4}+4x\)

Calculation:

\(\dfrac{5x-7}{3}+2=\dfrac{4x-3}{4}+4x\)

⇒ {(4x – 3)/4} + 4x – {(5x – 7)/3} = 2

⇒ 3(4x – 3) + 12 × 4x – 4(5x – 7) = 2 × 12      ----(Multiply with 12)

⇒ 12x – 9 + 48x – 20x + 28 = 24

⇒ 12x + 48x – 20x = 24 + 9 – 28

⇒ 40x = 5

⇒ x = 1/8

Now, (8x + 5) = 8/8 + 5 = 1 + 5 = 6

∴ The required value of the given expression is 6

Given:

x – 1/x = 4

Formula used:

(a – b)³ = a³ – b³ – 3ab(a – b)

Calculation:

Our given expression is x – 1/x = 4

Taking cube on both the sides,

(x – 1/x)³ = 4³

⇒ x³ – 1/x³ – 3x × 1/x(x – 1/x) = 64

⇒ x3 – 1/x3 – 3 × 4 = 64      ----(∵ x – 1/x = 4)

⇒ x3 – 1/x3 = 64 + 12

⇒ x3 – 1/x3 = 76

∴ The value of the given expression is 76

Given:

x(x - 5) = 7

⇒ x² – 5x = 7

Concept:

Cube both side of the given equation and then proceed.

Formula used:

(a - b)³ = a³ - b³ - 3ab(a - b)

Calculation:

∵ x² - 5x = 7

On cubing both sides, we get;

(x² – 5x)³ = 7³

⇒ x⁶ - 125x³ – [3 × (x²) × 5x(x² – 5x)] = 343

⇒ x⁶ - 125x³ - [3 × x² × (5x) × (7)] = 343

⇒ x⁶ - 125x³ - 105x³ = 343

⇒ x⁶ – 230x³ = 343      ------(1)

Putting the value of (1) in (x⁶ – 230x³ – 243)/4;

⇒ (343 – 243)/4

CALCULATION:

A:

Below given is the chart of cyclicity of the number:

13¹²³ can be written as: (13⁴)³⁰ × 13³

Unit digit of (13⁴)³⁰ is 1 while unit digit of 13³ is 7.

Hence, required unit digit = 7

B:

We know that:

(aⁿ – bⁿ) is divisible by (a + b) if n is even

So, (31²²⁴ – 19²²⁴) is completely divisible by (31 – 19) which is 12.

C:

72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²

48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3¹

64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶

Hence, HCF = 2³ = 8

Hence, only A and B are TRUE. 

Calculation:

l. x² + x – 2 = 0

⇒ x² + 2x – x – 2 = 0

⇒ x(x + 2) – 1(x + 2) = 0

⇒ (x – 1) (x + 2) = 0

⇒ x = 1, -2.

  ll. y² – 11y + 24 = 0

⇒ y² – 8y – 3y + 24 = 0

⇒ y(y – 8) – 3(y – 8) = 0

⇒ (y – 3) (y – 8) = 0

⇒ y = 3, 8.

∴ x < y.

Calculation:

I. x² + 29x + 208 = 0

⇒ x² + 16x + 13x + 208 = 0

⇒ x(x + 16) + 13(x + 16) = 0

⇒ (x + 16)(x + 13) = 0

⇒ x = -16 or x = -13

II. y² + 19y + 78 = 0 

⇒ y² +13y + 6y + 78 = 0

⇒ y(y + 13) + 6(y + 13) = 0

⇒ (y + 6)(y + 13) = 0

⇒ y = -6 or y = -13

Comparison between x and y (via Tabulation)

∴ x ≤ y

Given

X + Y = 60

And X ÷ Y = 5

Explanation

X/Y = 5

X = 5Y

X + Y = 60       ….(1)

Put X = 5Y in equation ( 1)

⇒  5 Y + Y = 60

⇒  6Y = 60

⇒  Y = 10

So, X is 50 i.e. (60 – 10)

Concept:

A number with unit digit 7 have repeating cycle of 7,9,3 and 1 after every fourth power.

Explanation:

Power series of 7 i.e. unit digit 7 power expansion has 7,9,3 and 1 and it is raised to power 753 i.e.

753 ÷ 4 = 188, remainder is 1

Given:

a² + b² = 82

ab = 9

Formula used:

(a + b)² = a² + b² + 2ab

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculation:

(a + b)3 = a3 + b3 + 3ab(a + b)

We need to find (a + b)

(a + b)2 = a2 + b2 + 2ab

⇒ (a + b)² = 82 + 2 × 9

⇒ (a + b)² = 100

⇒ (a + b) = 10

Now,

(a + b)3 = a3 + b3 + 3ab(a + b)

⇒ (10)³ = a³ + b³ + 3 × 9 × 10

⇒ 1000 = a³ + b³ + 270

⇒ a³ + b³ = 1000 - 270 = 730

∴ the value of a³ + b³ is 730

Given:

Nishu has chocolates = \(6\frac{1}{2}\) dozens

Nishu gave chocolates to Ravi = 1/6 of \(6\frac{1}{2}\) dozens

Nishu gave chocolates to Rahul = 1/13 of \(6\frac{1}{2}\) dozens

Half of the chocolates given to her friends = 1/2 of \(6\frac{1}{2}\) dozens

Concept used:

One dozen = 12

Calculations:

 \(6\frac{1}{2}\) dozens = (13/2) × 12

⇒ 13 × 6 = 78 chocolates 

The number of chocolates given to Ravi = (1/6) × \(6\frac{1}{2}\) dozens

⇒ 78 × 1/6 = 13 chocolates

The number of chocolates given to Rahul= (1/13) × \(6\frac{1}{2}\) dozens

⇒ 78 × 1/13 = 6 chocolates

The number of chocolates given to her friends = (1/2) of 78

⇒ (1/2) × 78 = 39

The number of chocolates given to friends, Ravi and Rahul = 39 + 6 + 13

⇒ 58

The number of chocolates left with her = Total chocolates - The number of chocolates given to other friends, Ravi and Rahul

⇒ 78 - 58 = 20

∴ The number of chocolates left with her is 20

Given:

x(3 – √7) = y(3 + √7) = 2

So y(3 + √7) = 2

And x(3 – √7) = 2

Concept:

From the given equations, calculate the value of ‘x’ and ‘y’.

Calculation:

∵ x(3 – √7) = 2

⇒ x = 2/(3 – √7)     ----(1)

∵ y(3 + √7) = 2

⇒ y = 2/(3 + √7)     ----(2)

From (1) and (2);

⇒ x + y = [2/(3 – √7)] + [2/(3 + √7)]

⇒ [2(3 + √7) + 2(3 – √7)]/[(3 – √7)(3 + √7)]

⇒ (6 + 2√7 + 6 – 2√7)/[(3)² – (√7)²]

⇒ 12/(9 – 7)

⇒ 12/2 = 6

∴ (x + y)/2 = 6/2 = 3

Given:

x² – 2x = 34

Formula used:

(x + y)² = x² + y² + 2ab

Calculation:

∵ x² – 2x = 34

Adding and subtracting 5x on LHS;

⇒ x² + 5x – 5x – 2x = 34

⇒ x² + 5x – 7x = 34

Subtracting 35 on both sides;

⇒ x² + 5x – 7x – 35 = 34 – 35

⇒ x(x + 5) – 7(x + 5) = (-1)

⇒ (x + 5)(x – 7) = (-1)

⇒ x – 7 = (-1)/(x + 5)

⇒ x + 1/(x + 5) = 7

Adding 5 both sides;

⇒ (x + 5) + 1/(x + 5) = 12

Squaring both sides;

⇒ (x + 5)² + 1/(x + 5)² + 2 = 144

Given:

a/b = 6/7

Calculation:

∵ (5/4) + (3b + 7)/(2a + 4)

⇒ (5/4) + [(3 × 7) + 7]/[(2 × 6) + 4]

⇒ (5/4) + (21 + 7)/(12 + 4)

⇒ (5/4) + 28/16

⇒ (5/4) + 7/4

⇒ (5 + 7)/4

⇒ 12/4 = 3

Given:

The number of women on the train is half the number of men. In Faridabad city, 10 men get off the train and five women enter the train.

Concept Used:

Concept of the linear equation of one variable

Calculation:

Let, the number of men in the train is 2x

Accordingly, the number of women on the train is x

In Faridabad city, 10 men get off the train and five women enter the train.

Now, the number of men in the train is (2x - 10) and the number of women in the train is (x + 5)

Accordingly,

(2x - 10) = (x + 5)

⇒ 2x - x = 5 + 10

⇒ x = 15

In the beginning, the number of women on the train was 15 and the number of men on the train was 15 × 2 = 30

At the beginning number of passengers in the train was (30 + 15) = 45

∴ In the beginning, 45 passengers entered the train.

Concept used:

(a – b)² = a² – 2ab + b² 

Calculations:

\(\sqrt{11 - 3 \sqrt{8}} = a + b \sqrt{2}\)

\(⇒ \sqrt{11 - 3 \sqrt{2 × 2 × 2}} = a + b \sqrt{2}\)

\(⇒ \sqrt{11 - 2 × 3 \sqrt{2}} = a + b \sqrt{2}\)

\(⇒ \sqrt{(3)^2 + (\sqrt2)^2 - 2 × 3 \sqrt{2}} = a + b \sqrt{2}\)

\(⇒ \sqrt{(3\;-\;√2)^2} = a + b \sqrt{2}\)

⇒ 3 – √2 = a + b√2

Compare a and b 

⇒ a = 3 

⇒ b = -1 

Value of (2a + 3b) = 2 × 3 + 3 × (-1) 

⇒ 6 – 3 = 3 

∴ Value of 2a + 3b is 3

Given:

- x - 4 = -3. We have to find the value of x

Concept Used:

concept of the linear equation of one variable.

Calculation:

- x - 4 = -3

⇒ - x = - 3 + 4

⇒  - x = 1

⇒ x = - 1

∴ The required value of x is - 1.

Concept used:

(a + b)² = a² + b² + 2ab 

Calculations:

2x2 – 7x + 5 = 0

⇒ 2x² + 5 = 7x

⇒ x² + 5/2 = 7x/2

⇒ x + 5/2x = 7/2 

Square of both side of equation

⇒ (x + 5/2x)² = (7/2)² 

⇒ x² + (5/2x)² + 2 × x × 5/2x = 49/4 

⇒ x² + 25/4x² + 5 = 49/4 

⇒ x² + 25/4x² = 49/4 – 5 

⇒ x² + 25/4x² = 29/4 

⇒ x² + 25/4x² = \(7 \frac{1}{4}\)

∴ The correct of answer is \(7 \frac{1}{4}\)

Given: 

I. x2 + 6x + 5 = 0

II. 8y2 - 22y - 21 = 0

Concept used:

Using quadratic equation.

Calculation:

From I,

x2 + 6x + 5 = 0

⇒ x² + 5x + x + 5 = 0

⇒ x(x + 5) + 1(x + 5) = 0

⇒ (x + 5)(x+1) = 0

⇒ x = - 5, -1

From II,

8y2 - 22y - 21 = 0

⇒ 8y² - 28y + 6y - 21 = 0

⇒ 4y(2y - 7) + 3(2y - 7) = 0

⇒ (2y - 7)(4y + 3) = 0

⇒ y = 7/2, - 3/4

Comparison between x and y (via Tabulation): 

∴ x < y.

Calculations:

From I,

x² – 6x + 8 = 0 

⇒ x² – 4x – 2x + 8 = 0 

⇒ (x – 4)(x – 2) = 0

⇒ x = 4, 2

From II,

y² – 11y + 30= 0 

⇒ y² – 6y – 5y + 30 = 0 

⇒ (y – 6)(y – 5) = 0

⇒ y = 6, 5

Comparison between x and y (via Tabulation):

∴ y > x

From I,

x² – 6x + 8 = 0 

⇒ x² – 4x – 2x + 8 = 0 

⇒ (x – 4)(x – 2) = 0

⇒ x = 4, 2

From II,

4y² – 24y + 32 = 0

⇒ 4y² - 16y - 8y + 32 = 0

⇒ (y - 4)(4y – 8) = 0

⇒ y = 4, 2

Comparison between x and y (via Tabulation):

∴ Relation between x and y can't be established.

x = y or the relationship between x and y cannot be established.

Calculation:

I: x² + 6x – 247 = 0

⇒ x² – 13x + 19x – 247 = 0

⇒ x × (x – 13) + 19× (x – 13) = 0

⇒ (x – 13) × (x + 19) = 0

⇒ x = (-19), 13

II: y² + 27y + 182 = 0

⇒ y² +14y + 13y + 182 = 0

⇒ y × (y + 14) + 13 × (y + 14) = 0

⇒ (y + 13) × (y + 14) = 0

⇒ y = -14, -13

Comparison between x and y (via Tabulation):

∴ x = y or the relationship between x and y cannot be established.

Given:

\(x + \frac{1}{x} = \;\surd 3\)

Calculation:

Taking cube both side of the equation,

\({x^3} + \;\frac{1}{{{x^3}}}\; + \;3x × \frac{1}{x}\left( {x\; + \;\frac{1}{x}} \right) = 3\sqrt 3 \)

⇒ \({x^3} + \;\frac{1}{{{x^3}}} + 3\sqrt 3 = 3\surd 3\)

⇒ \(\frac{{{x^6} + 1}}{{{x^3}}}\; = \;0\)

⇒ x⁶ + 1 = 0

⇒ x⁶ = - 1

x18 + x12 + x6 + 1

⇒ x¹²(x⁶ + 1) + (x6 + 1)

⇒ x¹² × 0 + 0

⇒ 0

∴ The value of x18 + x12 + x6 + 1 is 0.

Given:

The sum of the digits of a two-digit number = 12

The difference between the two digits of the number = 4

Calculations:

Let the two digits of the number be x and y respectively

⇒ x + y = 12      ----(i)

⇒ x – y = 4      ----(ii)

Adding (i) and (ii),

⇒ 2x = 16      

⇒ x = 8

Substituting the value of x in (i),

⇒ 8 + y = 12 

⇒ y = 4

The square root of the product of two digits = √(xy)

⇒ √(8 × 4)

⇒ √32

⇒ 4√2

∴ The square root of the product of the two digits of the number is 4√2    

Given: In an election, X and Y together got 19600 votes. Y won the election by a total number of 2760 votes.

Formula: Percentage increase/ decrease = [(new - actual)/actual] × 100%

Calculation: Let the total no. of votes of X and Y are x and y respectively.

According to question

x + y = 19600     ----(1)

y - x = 2760      ----(2)

On adding both the equations

2y = 22360

⇒ y = 11180

By putting value of y in any equation we get

x = 8420

∴ required answer is 8420.

I. 3x2 – 18x + 24 = 0

⇒ 3x2 – 12x – 6x + 24 = 0

⇒ 3x(x – 4) – 6(x – 4) = 0

⇒ (x – 4)(3x – 6) = 0

⇒ x = 4 or 2

II. 5y2 – 50y + 125 = 0

⇒ 5y2 – 25y – 25y – 125 = 0

⇒ 5y(y – 5) – 25(y – 5) = 0

⇒ (y – 5)(5y – 25) = 0

⇒ y = 5

Comparison between x and y (via Tabulation):

∴ x < y

I. 3x² + 11x + 6 = 0

⇒ 3x² + 9x + 2x + 6 = 0

⇒ 3x(x + 3) + 2(x + 3) = 0

⇒ (3x + 2)(x + 3)= 0

⇒ x = -2/3, -3

II. y² + 14y + 33 = 0

⇒ y² + 11y + 3y + 33 = 0

⇒ y(y + 11) + 3(y + 11) = 0

⇒ (y + 11)(y + 3) = 0

⇒ y = -11, -3

∴ x ≥ y

Given:

α and β are the roots of the equation 3x² – 11x + 6

Formula Used:

If ax² + bx + c = 0, then

Sum of the roots = -(b/a)

Product of the roots = (c/a)

(a – b)² = (a + b)² – 4 × a × b

Calculation:

For equation, 3x² – 11x + 6

a = 3, b = -11, and c = 6

The sum of the roots = -(b/a)

⇒ α + β = -(-11/3)

⇒ α + β = 11/3

The product of the roots = (c/a)

⇒ α × β = 6/3

⇒ α × β = 2

Now, (α - β)² = (α + β)² – 4 × α × β

⇒ (11/3)² – 4 × 2

⇒ (121/9) – 8

⇒ (121 – 72)/9

⇒ 49/9

∴ The value of (α - β)² is 49/9.

Calculation:

From I,

7x² + 32x – 15 = 0

⇒ 7x² + 35x – 3x – 15x = 0

⇒ 7x(x+5) – 3(x+5) = 0

⇒ (7x–3) (x+5) = 0

Taking,

⇒ 7x – 3 = 0 or x + 5 = 0

⇒ x = 3/7 or –5

From II,

4y²– 16y + 7 = 0

⇒ 4y²– 14y - 2y + 7 = 0

⇒ 2y(2y–1) –7(2y-1) = 0 

⇒ (2y–1) (2y–7) = 0

Taking, 

⇒ 2y – 1 = 0 or 2y – 7 = 0

⇒ y = 1/2 or y = 7/2

∴ Relation between x and y is, x < y. 

For unit digit divide the power of any integer by 4 and the remainder is the cycle for unit digit of the number

∴ Unit digit of 5⁴¹

⇒ 41/4 = 1 is the remainder i.e. cycle of 1 

⇒ 5¹ = 5

Unit digit of 7⁶⁹

⇒ 69/4 = 1 is the remainder 

⇒ 7¹ = 7

Unit digit of 3⁵⁷

⇒ 57/4 = 1 is the remainder 

⇒ 3¹ = 3

∴ Unit digit of 5⁴¹ × 7⁶⁹ × 3⁵⁷ = 5 × 7 × 3 = 105 or 5 is the unit digit

Given:

The difference between two digits of a two-digit number is 4, where the unit place number is great than the tenth-place number.

If digits are interchanged and the resulting number is added with the original number get 66 as result.

Concept Used:

1.) Linear equation of one variable

2.) In a two digit number if the unit place number is p and the tenth place number is q then the number should be (10q + p)

Calculation:

Let, the tenth-place number is x

The unit place number is great than the tenth-place number and difference between them is 4

Unit place number is (x + 4)

The number is {10x + (x + 4)} = (11x + 4)

When the digits interchange their position the number will become {10 × (x + 4) + x}

⇒ (11x + 40)

Accordingly,

(11x + 4) + (11x + 40) = 66

⇒ 22x + 44 = 66

⇒ 22x = 66 – 44

⇒ 22x = 22

⇒ x = 1

The tenth-place digit is 1 and the unit place digit is (1 + 4) = 5

∴ The original number is {(1 × 10) + 5} = 15

Identity used:

(a + b) × (a – b) = a² – b² 

Calculation:

(√2 + √3) × (√2 – √3) = (√2)² – (√3)² 

⇒ 2 – 3

⇒ –1

∴ The value of the given expression is –1

Given:

The sum of squares of three numbers is 138, while sum of their products taken two at a time is 131.

Concept used:

(x + y + z)² = x² + y² + z² + 2(xy +yz + zx)

Calculation:

Let three numbers be x, y, z.

The sum of squares of three numbers is 138

⇒ x² + y² + z² = 138

Sum of their products taken two at a time is 131

⇒ xy + yz + zx = 131

Applying the formulae 

⇒ (x + y + z)² = 138 + 2(131)

⇒ (x + y + z)² = 138 + 262

⇒ (x + y + z)² = 400

⇒ x + y + z = √400

⇒ x + y + z = 20

∴ The sum of three numbers is 20.

Given:

Factor of x² -3x -18 =?

Concept used:

To find the factor of an algebraic expression, put the equation value as zero

Detailed solution:

x² - 3x -18 = 0

⇒ x² - 6x + 3x - 18 = 0

⇒ x(x - 6) + 3(x - 6) = 0

⇒ ( x + 3)(x - 6) = 0

∴ The factor of the expression x² - 3x -18 is x - 6 and x + 3

Given:

We have to find the value of x from the given equation (x – 2)/(x + 3) = 2

Concept Used:

Concept of linear equation of single variable

Calculation:

(x – 2)/(x + 3) = 2

⇒ (x – 2) = 2 × (x + 3)

⇒ (x – 2) = 2x + 6

⇒ 2x – x = – 2 – 6

⇒ x = (–8)

∴ The value of x is (–8)

Given

The difference between two number = 3

Difference of their square = 69

Calculation

Let the greater number = x 

⇒ Smaller number = x - 3

From question 

⇒ x² - (x - 3)² = 69

⇒ x² - (x² - 2 × 3 × x + 3²) = 69

⇒ x² - x² + 6x - 9 = 69

⇒ 6x - 9 = 69 

⇒ 6x = 69 + 9

⇒ 6x = 78 

⇒ x =(78/6)

⇒ x = 13

⇒ Greater number = 13 and

⇒ Smaller number = 13 - 3

⇒ Smaller number = 10

∴ Greater number = 13 and smaller number = 10

Alternate method

Let small number = x 

⇒ Greater number = x + 3

From question 

⇒ (x + 3)² - x² = 69

⇒ (x² + 2 × x × 3 + 3²) - x² = 69

⇒ (x² + 6x + 9) - x² = 69

⇒ x² + 6x + 9 - x² = 69 

⇒ 6x + 9 = 69

⇒ 6x = 69 - 9

⇒ 6x = 60

⇒ x = (60/6)

⇒ x = 10 

⇒ Greater number = 10 + 3

⇒ Greater number = 13 

∴ Greater number = 13 and smaller number = 10

Given:

x – y = 13

xy = 25

Formula Used:

(x – y)² = x² + y² – 2xy

(x + y)² = x² + y² + 2xy

x² – y² = (x – y) × (x + y)

Calculation:

x – y = 13

Squaring both the sides

⇒ (x – y)² = 13²

⇒ x² + y² – 2xy = 169

⇒ x² + y² = 169 + 2 × 25

⇒ x² + y² = 219

(x + y)² = x² + y² + 2xy

⇒ (x + y)² = 219 + 2 × 25

⇒ (x + y) = √(269)

x² – y² = (x – y) (x + y)

⇒ x² – y² = 13√269

∴ The value of x² – y² is 13√269.  

Given

(x +(1/x)) = 2,

(x5 + x4 + x3 + x2 + x + 1) = ?

Calculation

⇒ (x +(1/x)) = 2

⇒ (x² + 1)/x = 2

⇒ x² + 1 = 2x 

⇒ x² - 2x + 1² = 0

⇒ (x - 1)² = 0 

⇒ x = 1 

Now, Put the value in the given equation

⇒ (x5 + x4 + x3 + x2 + x + 1) = (1⁵ + 1⁴ + 1³ + 1² + 1 + 1)

⇒ (x5 + x4 + x3 + x2 + x + 1) = 6

∴ (x5 + x4 + x3 + x2 + x + 1) = 6 

Alternate method

(x +(1/x)) = 2

In this type of question in which x +(1/x)) = 2,

You can directly put value of x = 1 

When you put x = 1 

⇒ x +(1/x)) = 2 

i.e R.H.S = L.HS

So, put the value of x = 1 in the equation

⇒ (x5 + x4 + x3 + x2 + x + 1) = 6

∴ (x5 + x4 + x3 + x2 + x + 1) = 6

Given:

x = √3 + √2

Formula used:

(a + b)² = a² + 2ab + b²

Calculation:

x = √3 + √2

⇒ 1/x = 1/(√3 + √2)

⇒ [1/(√3 + √2)] × [(√3 - √2)/(√3 - √2)]

⇒ (√3 - √2)

⇒ [x + (1/x)] = √3 + √2 + √3 - √2

⇒ [x + (1/x)] = 2√3

x² + (1/x)² = (2√3)² – 2

⇒ 10

x⁴ + (1/x)⁴ = [x + (1/x)]² – 2

⇒ 10² – 2

⇒ 100 – 2

⇒ 98

∴ x⁴ + (1/x)⁴ = 98.

Given:

x + y = 4

x³ + y³ = 12

Formula Used:

x² + y² = (x + y)² – 2xy

x³ + y³ = (x + y) (x² – xy + y²)

x⁴ + x⁴ = (x² + y²)² – 2x²y²

Calculation:

x³ + y³ = (x + y) (x² – xy + y²)

⇒ 12 = 4 × (x² – xy + y²)

⇒ x² + y² = 3 + xy

⇒ (x + y)² – 2xy = 3 + xy

⇒ 4² – 3 = 3xy
⇒ xy = (16 – 3)/3

⇒ xy = 13/3

x⁴ + x⁴ = (x² + y²)² – 2x²y²

⇒ (3 + xy)² – 2 × (13/3)²

⇒ [3 + (13/3)]² – 2 × (169/9)

⇒ [(9 + 13)/9]² – 338/9

⇒ (484/9) – 338/9

⇒ (484 – 338)/9

⇒ 146/9

∴ The value of x⁴ + y⁴ is 146/9.

Given:

X⁴ + 1/X⁴ = 194

Concept used:

(a + b)² = a² + b² + 2ab

Calculation:

X4 + 1/X4 = 194

⇒ (X²)² + 1/(X²)² = 194

⇒ (X2)2 + 1/(X2)² + 2 = 194 + 2                     [Adding 2 both sides]

⇒ (X2)2 + 1/(X2)2 + 2 × X² × 1/X² = 196

⇒ (X² + 1/X²)² = 196

⇒ X2 + 1/X² = √196

⇒ X2 + 1/X2 = 14

⇒ X2 + 1/X² + 2 = 14 + 2                                [Adding 2 both sides]

⇒ X2 + 1/X² + 2 × X × 1/X = 16

⇒ (X + 1/X)² = 16

⇒ X + 1/X = √16

⇒ X + 1/X = 4

X + 1/X + 2 = 4 + 2

⇒ X + 1/X + 2 = 6

∴ The value of X + 1/X + 2 is 6.

Calculation: 

As we know that,

x4 + x2y2 + y4 = (x² - xy +y²) (x² + xy + y²)

According to the question

x⁴ + x2y2 + y4 = 63, and

x2 + xy + y2 = 7          . . .      (1)

⇒ 63 = (x2 - xy +y2) × 7

⇒ (x2 - xy +y2) = 9        . . .   (2)

Add equation (1) and (2)

⇒ 2(x² + y²) = 16

⇒ x² + y² = 8

The value of (x² + y²) put in equation (1)

⇒ xy = 7 - 8 = -1

Now, 

⇒ (x/y + y/x)

⇒ (x² + y²)/xy

⇒ 8/-1

⇒ -8

∴ The required value is -8.

GIVEN:

 (10a3 + 4b3) : (11a3 - 15b3) = 7 : 5

CALCULATION:

 (10a3 + 4b3) : (11a3 - 15b3) = 7 : 5

⇒ 5 × (10a3 + 4b3)  = 7 × (11a3 - 15b3)

⇒ 50a³ + 20b³ = 77a³ - 105b³

⇒ 27a³ = 125b³

⇒ 3a = 5b

⇒ a : b = 5 : 3

⇒ (3a + 5b) : (9a - 2b) =?

⇒ (3 × 5 + 5 × 3) : (9 × 5 - 2 × 3) = ?

⇒ 30 : 39 = ?

⇒ 10 : 13 = ?

∴ (3 × 5 + 5 × 3) : (9 × 5 - 2 × 3) = 10 : 13

Calculation∶

I. 15x² - 14x + 3 = 0

⇒ 15x² - 5x - 9x + 3 = 0

⇒ 5x (3x - 1) - 3 (3x - 1) = 0
⇒ (3x - 1) (5x - 3) = 0

⇒ x = 1 / 3, 3 / 5

II. y² - 21y + 110 = 0

⇒ y² - 10y - 11y + 110 = 0

⇒ y (y - 10) - 11 (y - 10) = 0

⇒ (y - 10) (y - 11) = 0

⇒ y = 10, 11

Calculations:

From I,

2x² – 9x + 9 = 0

⇒ 2x² – 6x – 3x + 9 = 0

⇒ 2x(x – 3) - 3(x – 3) = 0

⇒ (x – 3)(2x - 3) = 0

Taking,

⇒ x – 3 = 0 or 2x - 3 = 0

⇒ x = 3 or x = 3/2

From II,

y² – 3y - 10 = 0

⇒ y² – 5y  + 2y – 10 = 0

⇒ y(y – 5) + 2(y – 5) = 0

⇒ (y – 5)(y + 2) = 0

Taking,

⇒ y – 5 = 0 or y + 2 = 0

⇒ y = 5 or y = –2

Comparison between x and y (via Tabulation):

∴ No relation in x and y.