## anonymous one year ago What reasoning and explanations can be used when solving radical equations and how do extraneous solutions arise from radical equations?

1. Michele_Laino

in order to solve a radical equation, we have to isolate the square rooth for example. then you have to square both sides, so you get an algebraic equation which doesn't contain any radical expressions Let's suppose to have this radical equation: $\sqrt {A\left( x \right)} = B\left( x \right)$ where A and B are two expressions, which depend on a variable x. Now we take the square of both sides, in so doing we get this new equation: $A\left( x \right) = B{\left( x \right)^2}$ Next, you have to keep in mind that the two equations above are not equivalent each other, in other words the second equation contains more solution with respect to the first one, we can write this: $A\left( x \right) = B{\left( x \right)^2} \Rightarrow A\left( x \right) - B{\left( x \right)^2} = 0$ so, we have: $A\left( x \right) - B{\left( x \right)^2} = \left\{ {\sqrt {A\left( x \right)} - B\left( x \right)} \right\}\left\{ {\sqrt {A\left( x \right)} + B\left( x \right)} \right\}=0$ In other words, the second equation contains the solution of our equation: $\sqrt {A\left( x \right)} = B\left( x \right)$ plus the solution of this one: ${\sqrt {A\left( x \right)} + B\left( x \right)}$ Those last solutions are the so called extraneous solutions

2. Michele_Laino

oops.. $\sqrt {A\left( x \right)} + B\left( x \right) = 0$

3. anonymous

Thank you so much!