If the whole expression 1 + 5x is under the square root sign:
1 + x = √(1 + 5x)
Remove the square root sign by squaring both sides:
( 1 + x )² = ( √(1 + 5x )²
1 + 2x + x² = 1 + 5x
Subtract 1 + 5x from both sides and rearranging:
x² - 3x = 0
x ( x - 3 ) = 0
So x = 0 or x = 3
But these are only possibilities! They must be checked!
1 + 0 = √(1 + 5·0) ---> 1 = √1 Yes!
1 + 3 = √(1 + 5·3) ---> 4 = √(16) = 4 Yes!
So, both 0 and 3 work.
1 + x = √(1 + 5x) square both sides
1 + 2x + x ^2 = 1 + 5x subtract 1 + 5x from both sides
-3x + x^2 = 0 rearrange
x^2 - 3x = 0 factor
x (x - 3) and setting each factor to 0, we get that x = 3 or x = 0
Make sure to check these answers....square root equations of this sort often give "extraneous" answers that will not work in the original problem......here, we're OK though !!!
If the whole expression 1 + 5x is under the square root sign:
1 + x = √(1 + 5x)
Remove the square root sign by squaring both sides:
( 1 + x )² = ( √(1 + 5x )²
1 + 2x + x² = 1 + 5x
Subtract 1 + 5x from both sides and rearranging:
x² - 3x = 0
x ( x - 3 ) = 0
So x = 0 or x = 3
But these are only possibilities! They must be checked!
1 + 0 = √(1 + 5·0) ---> 1 = √1 Yes!
1 + 3 = √(1 + 5·3) ---> 4 = √(16) = 4 Yes!
So, both 0 and 3 work.