Question

If x = (√(a+3b) + √(a-3b))/(√(a+3b) - √(a-3b)), prove that: 3bx^2 – 2ax + 3b = 0

Answer 1

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Answer 2

3bx² - 2ax + 3b = 0 (Proved)

Step-by-step explanation:

We are given that $x = \frac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}$ and we have to prove that 3bx² - 2ax + 3b = 0.

Now, $x = \frac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}$

⇒ $x = \frac{[\sqrt{a + 3b} + \sqrt{a - 3b}]^{2} }{(\sqrt{a + 3b}) ^{2} - (\sqrt{a - 3b}) ^{2} }$ {Rationalizing the denominator}

⇒  $x = \frac{a + 3b + a - 3b + 2\sqrt{a^{2} - (3b)^{2}}}{a + 3b - a + 3b}$

⇒ $6bx = 2a + 2\sqrt{a^{2} - 9b^{2}}$

⇒ $3bx - a = \sqrt{a^{2} - 9b^{2}}$

Now, squaring both sides we get,

⇒ 9b²x² - 6abx + a² = a² - 9b²

Dividing both sides with 3b, we get

⇒ 3bx² - 2ax + 3b = 0 (Proved)