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$$3\sqrt{3{x}^{2}+2x+36}+\sqrt{3{x}^{2}+2x-36}=40$$

How to solve? I see answer on other thread but want to know how to solve algebra.

I try $$y = 3{x}^{2}+2x$$, move over one radical, square both sides but not working.

Apr 15, 2020

#1
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Continuing from where you started:  [This was a great start!]

3sqrt(y + 36) + sqrt(y - 36)  =  40

3sqrt(y + 36)  =  40 - sqrt(y - 36)

Square both sides:

9(y + 36)  =  1600 - 80sqrt(y - 36) + (y - 36)

9y + 324  =  1564 + y - 80sqrt(y - 36)

8y - 1240  =  - 80sqrt(y - 36)

y - 15  =  -10sqrt(y - 36)

(155 - y)/10  =  sqrt(y - 36)

Square both sides:

(24025 - 310y + y2) / 100  =  y - 36

y2 - 310y + 24025  =  100y - 3600

y2 - 410y + 27625  =  0

y  =  [410 +/- sqrt(57600)] / 2

y  =  325  or  y  =  85

If  y= 325:                                                        If  y = 85

325  =  3x2 + 24                                               85  =  3x2 + 2x

3x2 + 2x - 325  =  0                                          3x2 + 2x - 85  =  0