MHB Radical Equation...Extraneous Roots

[mathdad]

When solving radical equations, we must check for extraneous roots. What are extraneous roots?

 

[Greg]

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.

 

[mathdad]

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.

I totally get it. By the way, your LaTex is blocked by your typing work of letters.

 

[HallsofIvy]

The same kind of thing can happen any time you multiply both sides of an equation by something involving the variable. For example, if we start with the very simple equation x= 4 and multiply on both sides by x- 3, we get x(x- 3)= 4(x- 3) which is the quadratic equation x^2- 7x+ 12= 0 which has solutions x= 3 and x= 4. Similarly, if you solve an equation involving algebraic fractions by multiplying both sides by the denominators, you may introduce "extraneous" roots that satisfy the new equation but not the original equation. "Squaring both sides of the equation" is equivalent to multiplying both sides of an equation by something.

 

[Greg]

By the way, your LaTex is blocked by your typing work of letters.

Thanks. I'll keep that in mind for future posts.

I've edited my post above.

 

[mathdad]

Cool.