Question 235956
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I'm guessing you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sqrt{x\ -\ 1}\ +\ 2\sqrt{x\ +\ 4}\ =\ 5]


Square both sides:  (use the FOIL process on the LHS)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(x\ -\ 1) + 8\sqrt{(x-1)(x+4)}\ + 4(x\ + 4)\ =\ 25]


Collect like terms and leave the radical term in the LHS and everything else in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8\sqrt{(x-1)(x+4)}\ =\ -8x\ +\ 13]


(Verification of that last step is left as an exercise for the student)


Square both sides again


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 64(x\ -\ 1)(x\ +\ 4)\ = 64x^2\ -\ 208x\ + 169]


Expand the binomial multiplication in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 64x^2\ +\ 192x\ -\ 256\ = 64x^2\ -\ 208x\ + 169]


And collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 400x\ =\ 425]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{17}{16}]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sqrt{\left(\frac{17}{16}\right)\ -\ 1}\ +\ 2\sqrt{\left(\frac{17}{16}\right)\ +\ 4}\ =^?\ 5]


You can do the arithmetic to see if it is right or not.


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