What's the deal with extraneous solutions when solving radical equations?

What's the deal with those extraneous solutions we get when solving radical equations? Why do we get two answers but sometimes have to throw out one or even both? In this post I share a video explaining through graphs the answers we get when solving radical equations.

What's the deal with those extraneous solutions we get when solving radical equations? Why do we get two answers but sometimes have to throw out one or even both? In this post I share a video explaining through graphs the answers we get when solving radical equations.



What's the deal with those extraneous solutions we get when solving radical equations? Why do we get two answers but sometimes have to throw out one or even both? In this post I share a video explaining through graphs the answers we get when solving radical equations.

It's easier to answer this question by asking another question first: What are radical functions? Radical functions are inverses to quadratic functions, except for one little thing... half of their graphs. Because if radicals were parabolas on their sides, 1-to-1 with every point on their quadratic graph inverse, they wouldn't be functions:

graph of a quadratic and its inverse, but the inverse is not a function

Above is a quadratic and the quadratic reflected over the y=x axis. The problem is that the red graph is not a function. It's a radical with its imaginary half showing. 

In order for the red graph to be a function, we have to throw out half of it. Now we have a quadratic and its radical inverse:

graph of a quadratic and its radical inverse

Now that we got that out of the way, let's look at an equation √(x + 7) = x + 5 on the grid. We'll graph it as y = √(x + 7) and y = x + 5 to solve. 

If we were to solve algebraically, we'd get two solutions: x = -6 and x = -3: 

graph of a radical equation we are solving through graphing

By plugging in, we'd find that x = -3 is a solution and x = -6 is an extraneous solution.

But what if that other half of our radical still existed?

graph of the radical equation we are solving through graphing and how the extraneous solution intersects the imaginary part of the radical

Look at that! If our radical was a parabola's 1-to-1 inverse, it would intersect y=x + 5 at our extraneous solution x = -6!

Here is a video:


In the video I mention a graphing radical cheat sheet, That can be found here:

Graphing radicals cheat sheet

Of course this doesn't explain all of the extraneous solutions we get when solving radical equations, like when a linear graph has no chance of ever intersecting the radical at all. But I have always found this to be a neat thing about radical equations when extraneous solutions come up!

See more radical functions activities

Scaffolded Math and Science blog


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