If
1
x + — = A
x
Then what is the value of
(1)
1
x – — = ?
x
Sol. (1)
1
x – — = ?
x
Now mind it that
(1)
1 1
x + — = A. Then Find x – — = ?
x x
The Formula is
A² (± Second Sign) 4
A² – 4
Example (1)
1 1
x + — = 2. Then Find x – — = ?
x x
Sol.1
Hire A = 2
The Formula is
√[A² (± Second Sign) 4]
= √(2² – 4)=> √(4 – 4) => √0 = 0
Example (2)
1 1
x + — = 3. Then Find x – — = ?
x x
Sol.2
Hire A = 3
The Formula is
√[A² (± Second Sign) 4]
= √(3² – 4)=> √(9 – 4) => √5
Example (3)
1 1
x + — = 4. Then Find x – — = ?
x x
Sol.3
Hire A = 4
The Formula is
√[A² (± Second Sign) 4]
= √(4² – 4)=> √(16 – 4) => √12
=> 2√3
Example (5)
1 1
x + — = 5. Then Find x – — = ?
x x
Sol.4
Hire A = 5
The Formula is
√[A² (± Second Sign) 4]
= √(5² – 4)=> √(25 – 4) => √21
(2)
1 1
x – — = A. Then Find x + — = ?
x x
The Formula is
A² (± Second Sign) 4
A² + 4
Example (1)
1 1
x – — = 2. Then Find x + — = ?
x x
Sol.1
Hire A = 2
The Formula is
√[A² (± Second Sign) 4]
= √(2² + 4)=> √(4 + 4) => √8 = 2√2
Example (2)
1 1
x – — = 3. Then Find x + — = ?
x x
Sol.2
Hire A = 3
The Formula is
√[A² (± Second Sign) 4]
= √(3² + 4)=> √(9 + 4) => √13
Example (3)
1 1
x – — = 4. Then Find x + — = ?
x x
Sol.3
Hire A = 4
The Formula is
√[A² (± Second Sign) 4]
= √(4² + 4)=> √(16 + 4) => √20
=> 2√5
Example (5)
1 1
x – — = 5. Then Find x + — = ?
x x
Sol.5
Hire A = 5
The Formula is
√[A² (± Second Sign) 4]
= √(5² + 4)=> √(25 + 4) => √29
Part B
1
x + — = A
x
Then what is the value of
(1)
1
x² + — = ?
x²
(2)
1
x³ + — = ?
x³
(3)
1
x⁴ + — = ?
x⁴
(4)
1
x⁵ + — = ?
x⁵
by X equals to a then x square
Sol.
(1)
1
x + — = A
x
Squaring both sides
1
(x + — )²= (A)²
x
by (a + b)² = (a² + b² +2a.b)
1 1
x² + — + 2× x × — = A²
x² x
1
x² + — + 2 = A²
x²
1
x² + — = A² – 2
x²
Sol.
(1)
1
x + — = A
x
Cubing both sides
1
(x + — )³= (A)³
x
(a + b)³ = (a + b) (a + b)²
1 1
x³ + — + 2× x × — = A²
x³ x
1
x² + — + 2 = A²
x²
1
x² + — = A² – 2
x²
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