Question

If x = 1/2 {√(a/b)-√(b/a)}, then prove that, 2a√(1+x^2) / x+√(1+x^2) = a+b

Answer 1

Answer:

2a(√x² + 1)(x + √x² + 1) = a + b.

Step-by-step explanation:

Hi,

Given x = 1/2 {√(a/b)-√(b/a)},

let t = b/a ,

(since square root is defined on t, t cannot be negative),

then x = 1/2{1/√t - √t]

x = 1/2{1 - t}/√t

2x√t = 1 - t

Squaring on both sides, we get

4x²t = 1 + t² - 2t

⇒ t² -2t(1 + 2x²) + 1 = 0

⇒ t = (2(1 + 2x²)  ± √4(1 + 2x²)² - 4)/2

But, t > 0

⇒ t = (1 + 2x²)  + √(1 + 2x²)² - 1

⇒ t = (1 + 2x²)  + √(1 + 4x² + 4x⁴ - 1

⇒ t = (1 + 2x²) + √4x² + 4x⁴

⇒ t = (1 + 2x²) + 2x√x² + 1

⇒ t + 1 = 1 +  (1 + 2x²) + 2x√x² + 1

⇒ t + 1 = 2( 1 + x² + x√x² + 1)

But t = b/a, so the above expression becomes

 b/a + 1 = 2( 1 + x² + x√x² + 1)

⇒ (a + b)/a = 2( 1 + x² + x√x² + 1)

⇒ a + b = 2a( 1 + x² + x√x² + 1)

Taking √x² + 1 as common, we get

a + b = 2a√x² + 1(x + √x² + 1).

Hence, 2a√x² + 1(x + √x² + 1) = a + b.

Hope, it helped !

Answer 2

2x=√(a/b)-√(b/a)

=(a-b)/√(ab)

x=(a-b)/2√(ab)

x^2=(a^2-2ab+b^2)/4ab

x^2+1=(a+b)^2/4ab

√(x^2+1)=(a+b)/2√(ab)

Now,

2a√(x^2+1)/x+√(x^2+1)

Put the value of x and √(x^2+1)

Therefore the solution is (a+b) Proved.