$$\frac{{\sqrt {3 + x} + \sqrt {3 - x} }}{{\sqrt {3 + x} - \sqrt {3 - x} }} = 2{\text{,}}$$ then x is equal to?
A. $$\frac{5}{{12}}$$
B. $$\frac{{12}}{5}$$
C. $$\frac{5}{7}$$
D. $$\frac{7}{5}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \frac{{\sqrt {3 + x} + \sqrt {3 - x} }}{{\sqrt {3 + x} - \sqrt {3 - x} }} = \frac{2}{1}{\text{ }} \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1} \cr & \left[ {\frac{{\text{A}}}{{\text{B}}} = \frac{{\text{C}}}{{\text{D}}}} \right] \cr & \left[ {\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = \frac{{{\text{C}} + {\text{D}}}}{{{\text{C}} - {\text{D}}}}} \right] \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = 3{\text{ }} \cr & {\text{Squaring both sides}} \cr & \Rightarrow \frac{{3 + x}}{{3 - x}} = 9 \cr & \Rightarrow 3 + x = 27 - 9x \cr & \Rightarrow 10x = 24 \cr & \Rightarrow x = \frac{{24}}{{10}} \cr & \Rightarrow x = \frac{{12}}{5} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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