Question

hi can you solve this

Answer 1

Answer:

Step-by-step explanation:

Given :

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

To prove :

$a^2b^2-2a^2x+b=0$

Proof :

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

By taking conjugate,

i.e. multiplying and dividing by $\sqrt{a^2+b^2}+\sqrt{a^2-b^2$

We get,

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

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

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

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

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

⇒ $x = \frac{a^2+\sqrt{a^4-b^4}}{b^2}$ ...(1)

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$a^2b^2-2a^2x+b=0$

⇒ $-2a^2x=-b-a^2b^2$ (or) $2a^2x = a^2b^2+b$

⇒ $x = \frac{a^2b^2+b}{2a^2}$ or $x = \frac{b^2}{2}+\frac{b}{2a^2}$  ....(2)

since LHS of (1) is equal to LHS of (2)  (LHS = Left Hand Side of equation)

but, RHS of (1) is equal to RHS of (2)  (RHS = Right Hand Side of equation).

It is incorrect,.