Checking solutions - textbook wrong about roots?

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The discussion centers on solving the equation sqrt(3x + 1) = x - 3. The solutions obtained by squaring both sides are x = 1 and x = 8. However, upon verification, x = 1 is deemed invalid because the left-hand side (LHS) evaluates to 2, while the right-hand side (RHS) evaluates to -2. The confusion arises from a common misconception regarding square roots, as the principal square root symbol (√) denotes only the positive root, confirming that the textbook is correct in stating that x = 1 is not a valid solution.

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Checking solutions -- textbook wrong about roots?

If I have the equation sqrt(3x + 1) = x - 3 and I need to solve for x, by squaring both sides then solving the resulting quadratic, I get the solutions x = 1, 8

However, since I squared the equation, I need to check if the solutions are valid. My calculus textbook says that x = 1 is not a valid solution as the LHS (sqrt(3(1) + 1)) = 2 and the RHS (1 - 3) = -2

However, the LHS also equals -2 as the square root of 4 is +/- 2.

So, is my textboox wrong? Or have I got the wrong idea somehow?
 
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BMW said:
If I have the equation sqrt(3x + 1) = x - 3 and I need to solve for x, by squaring both sides then solving the resulting quadratic, I get the solutions x = 1, 8

However, since I squared the equation, I need to check if the solutions are valid. My calculus textbook says that x = 1 is not a valid solution as the LHS (sqrt(3(1) + 1)) = 2 and the RHS (1 - 3) = -2

However, the LHS also equals -2 as the square root of 4 is +/- 2.
No, √4 = 2. It is a common misconception to think that, for example, √(25) = ± 5. The symbol √x, where x is a positive number, represents the positive square root.
BMW said:
So, is my textboox wrong? Or have I got the wrong idea somehow?
 

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