Why some algebraic solutions are mathematically impossible
In short: When solving radical equations, you can sometimes arrive at a value that doesn't actually work in the original problem. This video explores how to identify these extraneous solutions using a specific example.
The tutorial focuses on finding a specific constant, d, for which the equation √(3x+25) = d+2x yields an extraneous solution at x=-3. By understanding why these phantom answers appear, students can better grasp the constraints of radical expressions.
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Sources
- Summary by Intellect · source: Extraneous solutions of radical equations (example 2) | High School Math | Khan Academy
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