QQuestion
Using componendo and dividendo, solve for x:
\frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}} = 3
✓Answer
Given:
\frac{\sqrt{2x+2}+\sqrt{2x-1}}{\sqrt{2x+2}-\sqrt{2x-1}}=3Applying componendo and dividendo:
\frac{(\sqrt{2x+2}+\sqrt{2x-1})+(\sqrt{2x+2}-\sqrt{2x-1})}{(\sqrt{2x+2}+\sqrt{2x-1})-(\sqrt{2x+2}-\sqrt{2x-1})}=\frac{3+1}{3-1}⇒ \frac{2\sqrt{2x+2}}{2\sqrt{2x-1}}=\frac{4}{2}
⇒ \frac{\sqrt{2x+2}}{\sqrt{2x-1}}=2
⇒ √(2x + 2) = 2√(2x − 1)
Squaring both sides:
⇒ 2x + 2 = 4(2x − 1)
⇒ 2x + 2 = 8x − 4
⇒ 8x − 2x = 2 + 4
⇒ 6x = 6
⇒ x = 1
Therefore, x = 1.
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