Question

please solve this 2 questions ​

Answer 1

Given :-

• x = 1/(3 - 2√2)

To Find :-

• (x² + 1/x²) = ?

Solution :-

→ x = 1/(3 - 2√2)

→ 1/x = (3 - 2√2) ---------- Equation (1).

And,

→ x = 1/(3 - 2√2)

Rationalising RHS side we get

→ x = 1/(3 - 2√2) * [ (3 + 2√2) / (3 + 2√2) ]

→ x = (3 + 2√2) / (3² - (2√2)²)

→ x = (3 + 2√2) / (9 - 8)

→ x = (3 + 2√2) ------------- Equation (2).

So,

→ (x + 1/x) = (3 - √2) + (3 + 2√2)

→ (x + 1/x) = 6

Squaring both sides now,

→ (x + 1/x)² = 6²

→ x² + 1/x² + 2 * x * 1/x = 36

→ (x² + 1/x²) = 36 - 2

→ (x² + 1/x²) = 34 (Ans.)

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Solution (2) :-

Factorise :-

→ x³ - 3x² - 9x - 5

→ x³ + x² - 4x² - 4x - 5x - 5

→ x²(x+1) - 4x(x+1) - 5(x+1)

→ (x + 1) (x² - 4x - 5)

→ (x + 1) (x² - 5x + x - 5)

→ (x + 1) [ x(x - 5) + 1(x - 5) ]

→ (x + 1) [ (x - 5)(x+1) ]

→ (x + 1)(x+1)(x - 5) (Ans.)

Answer 2

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Question 27.

$\huge\tt{GIVEN:}$

• x = 1/3-2√2

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$\huge\tt{TO~FIND:}$

• x²+ 1/x²

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➩ x = 1/(3 - 2√2)

➩ 1/x = (3 - 2√2) _______(EQ.1)

➩x = 1/(3 - 2√2)

➩x = 1/(3 - 2√2) × [ (3 + 2√2) / (3 + 2√2) ]

➩x = (3 + 2√2) / (3² - (2√2)²)

➩x = (3 + 2√2) / (9 - 8)

➩x = (3 + 2√2) _________(EQ.2)

So,

➩ (x + 1/x) = (3 - √2) + (3 + 2√2)

➩ (x + 1/x) = 6

➩ (x + 1/x)² = 6²

➩ x² + 1/x² + 2 × x × 1/x = 36

➩ (x² + 1/x²) = 36 - 2

➩ (x² + 1/x²) = 34

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Question 28.

$\huge\tt{GIVEN:}$

• x³ - 3x² - 9x - 5

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$\huge\tt{TO~FIND:}$

• Factorising it

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$\huge\tt{SOLUTION:}$

➠x³ - 3x² - 9x - 5

➩ x³ + x² - 4x² - 4x - 5x - 5

➩ x²(x+1) - 4x(x+1) -5(x+1)

➩ (x+1)(x² - 4x - 5)

➩ (x+1)(x² - 5x + x - 5)

➩(x+1)[x(x-5) + 1(x-5)]

➩ (x+1)[(x-5)(x+1)]

➩ (x+1)(x+1)(x-5)

➩(x+1)²(x-5)

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