solve-for-x-2-x-2-x-2-x-2-x-3-

Question Number 62489 by Tawa1 last updated on 21/Jun/19

solve for x : \frac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = 3

Answered by som(math1967) last updated on 22/Jun/19

\frac{\sqrt{2 - x} + \sqrt{2 + x} + \sqrt{2 - x} - \sqrt{2 + x}}{\sqrt{2 - x} + \sqrt{2 + x} - \sqrt{2 - x} + \sqrt{2 + x}} = \frac{3 + 1}{3 - 1} ★

\frac{2 \sqrt{2 - x}}{2 \sqrt{2 + x}} = \frac{4}{2}

\frac{2 - x}{2 + x} = 4 ★ ★

8 + 4 x = 2 - x \Rightarrow x = - \frac{6}{5} a n s

★ u s i n g c o m p o n e n d o a n d d i v i d e n d o

★ ★ s q u a r i n g b o t h s i d e

Commented by MJS last updated on 22/Jun/19

you should make your first step clearer .

in fact there are a few steps :

\frac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = 3

\sqrt{2 - x} + \sqrt{2 + x} = 3 \sqrt{2 - x} - 3 \sqrt{2 + x}

4 \sqrt{2 + x} = 2 \sqrt{2 - x}

2 \sqrt{2 + x} = \sqrt{2 - x}

4 (2 + x) = 2 - x

6 + 5 x = 0

x = - \frac{6}{5}

Commented by Tawa1 last updated on 22/Jun/19

God bless you

Commented by peter frank last updated on 22/Jun/19

n i c e t o o

Answered by MJS last updated on 21/Jun/19

defined for

2 - x ⩾ 0 \land 2 + x ⩾ 0 \land \sqrt{2 - x} \neq \sqrt{2 + x}

\Rightarrow - 2 ⩽ x ⩽ 2 \land x \neq 0

\frac{{(\sqrt{2 - x} + \sqrt{2 + x})}^{2}}{(\sqrt{2 - x} - \sqrt{2 + x}) (\sqrt{2 - x} + \sqrt{2 + x})} = 3

- \frac{2 + \sqrt{2 - x} \sqrt{2 + x}}{x} = 3

\sqrt{2 - x} \sqrt{2 + x} = - 3 x - 2

4 - x^{2} = 9 x^{2} + 12 x + 4

10 x^{2} + 12 x = 0

x (5 x + 6) = 0

x_{1} = 0 [not valid]

x_{2} = - \frac{6}{5}

Commented by Tawa1 last updated on 21/Jun/19

Good bless you sir

Commented by MJS last updated on 22/Jun/19

the path of s o m (m a t h 67) is better, I {didn}^{'} t

realize this possibility at first (always looking

for trouble, my thinking is too complicated

sometimes)

Commented by Tawa1 last updated on 22/Jun/19

Hahahaha

Commented by peter frank last updated on 22/Jun/19

t h a n k y o u

Answered by Rasheed.Sindhi last updated on 22/Jun/19

A n O ther W ay

\frac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = 3

\Rightarrow \frac{\frac{\sqrt{2 - x}}{\sqrt{2 + x}} + 1}{\frac{\sqrt{2 - x}}{\sqrt{2 + x}} - 1} = 3

\Rightarrow \frac{\frac{\sqrt{2 - x}}{\sqrt{2 + x}} + 1}{\frac{\sqrt{2 - x}}{\sqrt{2 + x}} - 1} = \frac{3}{1} = \frac{2 + 1}{2 - 1}

\frac{\sqrt{2 - x}}{\sqrt{2 + x}} = 2

\frac{2 - x}{2 + x} = 4

8 + 4 x = 2 - x

5 x = - 6

x = - 6 / 5

Commented by Tawa1 last updated on 22/Jun/19

God bless you sir

Commented by peter frank last updated on 22/Jun/19

v e r y n i c e

Commented by Rasheed.Sindhi last updated on 22/Jun/19

⊤ ∦ \forall \cap k ≶!