\(3\sqrt{3{x}^{2}+2x+36}+\sqrt{3{x}^{2}+2x-36}=40\)
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.
+23246 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
Use the quadratic formula:
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
Use the quadratic formula:
x = [ -1 +/- sqrt(3904) ] / 6 x = [ -2 +/- sqrt(1024) ] / 6
These answers don't check! Either x = 5 or x = -5 2/3
