how would you solve 1-3x=2rootx by hand?

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answers:
Jeffrey K say: Square both sides
(1-3x)^2 = 4x
1-6x+9x^2 = 4x
You can solve the quadratic equation from here.
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lenpol7 say: Square up to remove the 'squre root'.
( 1 - 3x)^2 = 4x
1 - 6x + 9x^2 = 4x
9x^2 - 10x + 1 = 0
Factor
(9x - 1)(x - 1) = 0
9x - 1 = 0
x = 1/9
&
x - 1 = 0
x = 1
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la console say: 1 - 3x = 2√x → you square both sides

(1 - 3x)² = (2√x)²

1 - 6x + 9x² = 4x

9x² - 10x + 1 = 0

9x² - (9x + x) + 1 = 0

9x² - 9x - x + 1 = 0

(9x² - 9x) - (x - 1) = 0

9x.(x - 1) - (x - 1) = 0

(x - 1).(9x - 1) = 0


First case: (x - 1) = 0 → x - 1 = 0 → x = 1

Second case: (9x - 1) = 0 → 9x - 1 = 0 → 9x = 1 → x = 1/9
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khalil say: 1 - 3x = 2√x
square it
(1 - 3x )² = 4x
1+ 9x² - 6x - 4x = 0
9x² - 10x + 1 = 0
x = 5 ± √(5² - 9) / 9
x = 1/9 or 1
test
x = 1/ 9 → 1 - 3(1/9) = 2√(1/9) ...accepted
x = 1 → 1 - 3(1) = 2√1 ....not accepted
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