Question

If x=√2a+1+√2a-1/√2a+1-√2a-1,then prove that x^2-4ax+1=0.

Plz....Plz....Please answer it....Plz....

Answer 1

find this solution............

Answer 2

x² - 4ax + 1 = 0 (Proved)

Step-by-step explanation:

We are given that, $x = \frac{\sqrt{2a + 1} + \sqrt{2a - 1} }{\sqrt{2a + 1} - \sqrt{2a - 1}}$ and we have to prove that x² - 4ax + 1 = 0.

Now, $x = \frac{\sqrt{2a + 1} + \sqrt{2a - 1} }{\sqrt{2a + 1} - \sqrt{2a - 1}}$

Rationalizing the denominator we get,

$x = \frac{(\sqrt{2a + 1} + \sqrt{2a - 1})^{2}}{2a + 1 - 2a + 1}$

⇒ $x = \frac{2a + 1 + 2a - 1 + 2 \sqrt{(2a + 1)(2a - 1)}}{2}$

⇒ $x = 2a + \sqrt{4a^{2} - 1}$

⇒ $x - 2a = \sqrt{4a^{2} - 1}$

Now, squaring both sides we get,

x² - 4ax + 4a² = 4a² - 1

⇒ x² - 4ax + 1 = 0 (Proved)