Question

If x = (root a+2b + root a-2b)/( root a+2b - root a-2b)

show that bx^2 - ax + b = 0

Answer 1

Heya mate....

↪️Two attachments are above which will help uh in ur question.

✍️The question is solved by using Componendo and Dividendo.

By applying that method we get the answer.

Hope u will understand.. ☺️✌️

Answer 2

Answer:

Step-by-step explanation:

The given expression is x= $\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}$

Now we will simplify the denominator of the given fraction

x = $\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\times \frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}+\sqrt{a-2b}}$

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

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

x = $\frac{a+2b+a-2b+2\sqrt{a^{2}-4b^{2}}}{4b}$

x = $\frac{2a+2\sqrt{a^{2}-4b^{2}}}{4b}$

x = $\frac{a+\sqrt{a^{2}-4b^{2}}}{2b}$

2bx = a + √(a² - 4b²)

(2bx - a)² = a² - 4b²

4b²x² + a² - 4abx = a² - 4b²

4b²x² -4abx = -4b²

bx² - ax = -b

bx² - ax + b = 0

Learn more to solve the radical fractions from https://brainly.in/question/7989267