Math · College algebra · Worked example
Solve a radical equation and check the roots
Solve √(x + 7) + 5 = x.
Isolate the square root
Subtract 5 from both sides. The root now stands alone, and since a principal square root is never negative, a solution must also make x − 5 nonnegative.
Square both sides
The left side becomes x + 7. The right side is the square of a binomial, (x − 5)² = x² − 10x + 25.
Solve the quadratic
Move every term to one side and factor. The squared equation has two candidates, x = 2 and x = 9.
Check x = 9 in the original equation
√(9 + 7) + 5 = 4 + 5 = 9, which equals the right side, 9. So x = 9 is a solution.
Check x = 2 in the original equation
√(2 + 7) + 5 = 3 + 5 = 8, but the right side is 2. So x = 2 is extraneous: it satisfies the squared equation only because √9 = 3 and x − 5 = −3 have equal squares.
See why the graph agrees
The curve y = √(x + 7) + 5 starts at (−7, 5) and rises slowly, while the line y = x rises with slope 1. They meet once, at (9, 9). At x = 2 the curve is at height 8 and the line at height 2, so they do not meet there.
Result
x = 9. The candidate x = 2 is extraneous.
Your turn
Solve √(3x + 1) = x − 1.
Show the answer and explanation
x = 5 (x = 0 is extraneous).
Squaring gives 3x + 1 = x² − 2x + 1, so x² − 5x = 0 and x(x − 5) = 0. Check: x = 5 gives √16 = 4 and 5 − 1 = 4, which works; x = 0 gives √1 = 1 but 0 − 1 = −1, which fails.
Keep exploring
Open the steps in Math and write x = 2 as the last line instead: the checker reports that it does not satisfy the original equation.
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