Question

If x = `(sqrt(b^2 + ab) + sqrt(b^2 - ab))/(sqrt(b^2 + ab) - sqrt(b^2 - ab))` show that ax² − 2bx + a = 0.

Answer 1

We are given that,

x = `(sqrt(b^2 + ab) + sqrt(b^2 - ab))/(sqrt(b^2 + ab) - sqrt(b^2 - ab))`

Divide both the numerator and the denominator inside the square roots by b:

⇒ x = `(sqrt(b(b + a)) + sqrt(b(b - a)))/(sqrt(b(b + a)) - sqrt(b(b - a))) = (sqrtb sqrt(b + a) + sqrtb sqrt(b - a))/(sqrtb sqrt(b + a) - sqrtb sqrt(b - a))`

Factor out `sqrtb` from the numerator and denominator:

⇒ x = `(sqrtb (sqrt (b + a) + sqrt(b - a)))/(sqrtb (sqrt (b + a) - sqrt(b - a))) = (sqrt(b + a) + sqrt(b - a))/(sqrt(b + a) - sqrt(b - a))`

Apply componendo and dividendo:

⇒ `(x + 1)/(x - 1) = ((sqrt(b + a) + sqrt(b - a)) + (sqrt(b + a) - sqrt(b - a)))/((sqrt(b + a) + sqrt(b - a)) - (sqrt(b + a) - sqrt(b - a)))`

⇒ `(x + 1)/(x - 1) = (2sqrt(b + a))/(2sqrt(b - a))`

⇒ `(x + 1)/(x - 1) = sqrt((b + a)/(b - a))`

Square both sides:

⇒ `((x + 1)/(x - 1))^2 = (b + a)/(b - a)`

⇒ `(x^2 + 2x + 1)/(x^2 - 2x + 1) = (b + a)/(b - a)`

Apply componendo and dividendo again:

⇒ `((x^2 + 2x + 1) + (x^2 - 2x + 1))/((x^2 + 2x + 1) - (x^2 - 2x + 1)) = ((b + a) + (b - a))/((b + a) - (b - a))`

⇒ `(2x^2 + 2)/(4x) = (2b)/(2a)`

⇒ `(x^2 + 1)(2x) = b/a`

⇒ a(x² + 1) = 2bx

⇒ ax² + a = 2bx

Rearranging the terms yields the required equation:

⇒ ax² − 2bx + a = 0