Question 188628
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  sqrt{2x+1}-sqrt{x+1} = 2]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (2x+1) - 2 sqrt{(2x+1)(x+1)} + (x + 1) = 4]


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  - 2 sqrt{(2x+1)(x+1)} = -3x + 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  sqrt{(2x+1)(x+1)} = \frac{3x}{2} - 1]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (2x+1)(x+1) = \frac{9x^2}{4} - 3x + 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2x^2 + 3x + 1 = \frac{9x^2}{4} - 3x + 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  8x^2 + 12x + 4 = 9x^2 - 12x + 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -x^2 + 24x = 0 \ \ \Rightarrow\ \ x^2 - 24x = 0 \ \ \Rightarrow\ \ x(x - 24) = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  sqrt{2(0)+1}-sqrt{(0)+1} = 1 - 1 = 0 \neq 2]


Root is extraneous


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  sqrt{2(24)+1}-sqrt{(24)+1} = sqrt{49} - sqrt{25} = 7 - 5 = 2]


Answer checks.


This problem brings to mind a very interesting brain teaser.  The number 48 is very special.  If you add 1, you get 49 which is a perfect square.  If you divide it by 2 and then add 1, you get another perfect square (48/2 = 24, 24 + 1 = 25).  Find three other integers with the same property.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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