Nice problem to consider: real roots of sqrt(x) = bx + c, where b>0.

[Steven G]

Consider the equation sqrt(x) = bx + c, where b>0.

Under which conditions will we, if ever, get 2 reals roots, 1 real and 1 erroneous roots,...?

 

[Dr.Peterson]

Consider the equation sqrt(x) = bx + c, where b>0.

Under which conditions will we, if ever, get 2 reals roots, 1 real and 1 erroneous roots,...?

What are your thoughts?

The big question is, what do you mean by "1 real and 1 erroneous roots"? An erroneous root would be one that you got by mistake - not a root at all! And an extraneous root is a result of a particular method of solution, not a property of the equation itself. (It, too, is not really a root at all.) But I suppose you are thinking of the fact that the line on the RHS might intersect the parabola of which the LHS is part, in 2 points, or 1, or none; and one or both of the intersections might be below the x-axis, which would naturally be thought of as extraneous roots of the given equation in some sense.

Guided by the graph, I would consider, for any given b, the value of c for which the line would be tangent to the parabola, and what the sign of c is. All this can probably be done with or without calculus. Letting b be negative would add in some interesting complexity.

 

[apple2357]

If the product of b and c is 0.25, you appear to get a tangent ( b>0, c>0)

 

[Dr.Peterson]

If the product of b and c is 0.25, you appear to get a tangent ( b>0, c>0)

Correct. How did you determine that? And do you say "appear to get", rather than "do", because you didn't use a method that is convincing? I'd like to see your method.

I can see ways to do this by calculus, or by solving the radical equation. There may be other methods I haven't thought of!

 

[apple2357]

Correct. How did you determine that? And do you say "appear to get", rather than "do", because you didn't use a method that is convincing? I'd like to see your method.

I can see ways to do this by calculus, or by solving the radical equation. There may be other methods I haven't thought of!

Just solving the equation and thinking about the discriminant of the quadratic. I then played with it on a graphing software to check further.
Do you need more detail?

 

[Dr.Peterson]

Just solving the equation and thinking about the discriminant of the quadratic. I then played with it on a graphing software to check further.
Do you need more detail?

That would be the algebraic approach I had in mind. No need for details unless someone wants help with it! As it is, it's a nice teaser for them.