Solve for x and check any extraneous roots : root 5x-1=1 root x+2

√(5x - 1) = 1

[√(5x - 1)]

5x - 1 = 1

5x = 2

[√(5x - 1)]

^{2}= 1^{2}5x - 1 = 1

5x = 2

*x = 2/5*Let's check solution in the original equation

√(5 * 2/5 - 1) = 1

√(2 - 1) = 1

√1 = 1

1 = 1

√(5 * 2/5 - 1) = 1

√(2 - 1) = 1

√1 = 1

1 = 1

*Thus, x = 2/5 is not an extraneous root.*
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