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mathdad
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When solving radical equations, we must check for extraneous roots. What are extraneous roots?
greg1313 said:Consider $\sqrt{6-2\sqrt5}=y$. Both $1-\sqrt5$ and $\sqrt5-1$ give $6-2\sqrt5$ when squared, but only $\sqrt5-1$ can be a root as $\sqrt{6-2\sqrt5}$ is positive. The "root" $1-\sqrt5$ is extraneous.
An extraneous root is a root induced by some mathematical operation in the method of solving that does not satisfy the original equation.
RTCNTC said:By the way, your LaTex is blocked by your typing work of letters.
A radical equation is an equation that contains a radical expression, such as a square root or cube root, with a variable in the radicand (the number under the radical symbol).
An extraneous root in a radical equation is a solution that does not satisfy the original equation. This can occur when the equation is manipulated in a way that introduces additional solutions that are not valid for the original equation.
To solve a radical equation, you need to isolate the radical expression and then raise both sides of the equation to the appropriate power to eliminate the radical. This will typically result in two solutions, so it's important to check for extraneous roots.
One common strategy for dealing with extraneous roots is to check each solution in the original equation to determine if it is a valid solution. Another strategy is to restrict the domain of the equation to only include values that will not result in extraneous roots.
Yes, a radical equation can have multiple extraneous roots. This can occur when there are multiple radical expressions in the equation or when the equation is manipulated in a way that introduces multiple extraneous roots.