Question

Pls answer the below question answer question 2 and 3

Answer 1

The answer is given below :

2.

Given,

x = √(m+n) + √(m-n)}/{√(m+n) - √(m-n)}

Now, we rationalise the denominator by multiplying both the numerator and the denominator by {√(m+n) + √(m-n)}, we get

x = [{√(m+n) + √(m-n)}{√(m+n) + √(m-n)}]/
{√(m+n) - √(m-n)}{√(m+n) + √(m-n)}

= [√(m+n) + √(m-n)]²/[{√(m+n)}² - {√(m-n)}²],
since (a+b)(a+b) = (a+b)²
and (a+b)(a-b) = a²-b²

= [(m+n) + 2√(m²-n²) + (m-n)]/[(m+n) - (m-n)],
since (a+b)² = a² + 2ab + b²

= [2m+2√(m²-n²)]/(m+n-m+n)

= [2m+2√(m²-n²)]/2n

= [m+√(m²-n²)]/n

⇒ nx = m+√(m²-n²)

⇒ nx - m = √(m²-n²)

Now, squaring both sides, we get

(nx - m)² = m² - n²

⇒ n²x² - 2mnx + m² = m² - n²

⇒ n²x² - 2mnx + n² = 0

⇒ nx² - 2mx + n = 0,
dividing both sides by n, n≠0.
(Ans.)

3.

Now,

9 + 4√5

= 5 + 4√5 + 4

= (√5)² + (2 × √5 × 2) + 2²

= (√5 + 2)²

So, √(9 + 4√5) = √5 + 2

Hence,

x = 1/√(9+4√5)

= 1/(√5 + 2)

= (√5 - 2){(√5 + 2)(√5 - 2),
by rationalising the denominator

= (√5 - 2)/(5 - 4),
since (a+b)(a-b) = a²-b²

= (√5 - 2)/1

= √5 - 2

Given that,

p(x) = x³ + 7x² + 16x + 7

So, p(√5 - 2)

= (√5 - 2)³ + 7(√5 - 2)² + 16(√5 - 2) + 7

= (√5)³ - {3 × (√5)² × 2} + (3 × √5 × 2²) - 2³
+ 7{(√5)² - (2 × √5 × 2) + 2²} + 16(√5 - 2) + 7,
using (a-b)³ = a³ - 3a²b + 3ab² - b³

= (5√5 - 30 + 12√5 - 8) + 7(5 - 4√5 + 4)
+ 16(√5 - 2) + 7

= 17√5 - 38 + 7(9 - 4√5) + 16(√5 - 2) + 7

= 17√5 - 38 + 63 - 28√5 + 16√5 + 32 + 7

= (17 - 28 + 16)√5 + (- 38 + 63 + 32 + 7)

= 5√5 + 0

= 5√5

So, option (4) is correct.

Thank you for your question.

Answer 2

the correct answer is the option no.(4)

5root5

hope it helps