Why is one root rejected when solving equations with multiple solutions?

[zorro]

When we solve to equations say y² = 4x and x² + y² = 9/4 for x, we get two values -

x=0.5 and x=-4.5

The equations represent a parabola and a circle resp. with x=either 0.5 or -4.5 as the abcissae of intersection. I don't understand why do we reject one value which is -4.5 in this case. By drawing a figure we can find out why but I am interested in knowing the reason behind -4.5 as one of the solutions. Why -4.5 is not a point of intersection? Why do we get this value if it is not a solution?

 

[zetafunction]

if you put x=-4.5 inside the first equation

$$y^{2}=-18$$ this makes NO sense for real numbers

 

[muppet]

Hi Abdul,
It seems to me that if you assume y is a real number, then the first equation constrains x to be positive. I'm not algebra-savvy enough to give you a precise answer to the question of why -4.5 appears, except to note that any quadratic will give you two (possibly complex) roots (counted with multiplicity), so some second solution must exist. I can't think of a good reason to pin it down more precisely than that though; I'll have a think...

 

[Hurkyl]

Why -4.5 is not a point of intersection?

Because it's not a point at all. 0.5 isn't a point of intersection either.

Why do we get this value if it is not a solution?

I suspect you are holding a widely held misunderstanding about solving problems.


When you (correctly) do these algebraic manipulations, you are making the assertion:

Any solution to the original problem is also a solution to the new problem​

which is a very different statement than

Any solution to the new problem is also a solution to the original problem​

or even

The new problem and the original problem have the same set of solutions​

 

[zorro]

Because it's not a point at all. 0.5 isn't a point of intersection either.

What are these values then?

I suspect you are holding a widely held misunderstanding about solving problems.

When you (correctly) do these algebraic manipulations, you are making the assertion:

Any solution to the original problem is also a solution to the new problem​

which is a very different statement than

Any solution to the new problem is also a solution to the original problem​

or even

The new problem and the original problem have the same set of solutions​

err I don't understand what are you trying to convey. Can you please explain it more clearly?

 

[zorro]

if you put x=-4.5 inside the first equation

$$y^{2}=-18$$ this makes NO sense for real numbers

If it doesnot make sense why do we get that value?

 

[Hurkyl]

What are these values then?

A point of intersection has two coordinates. (and the coordinates satisfy the original system of equations)

err I don't understand what are you trying to convey. Can you please explain it more clearly?

What part don't you understand? Do you not understand what the English sentence

Any solution to the original problem is also a solution to the new problem​

means? Do you not understand what I mean by "new problem" and "original problem"? Something else?

 

[CompuChip]

Here's a simple example.
Consider the simple equation x = 2.
Of course, the equation already states the solution: for x = 2 it is true.
But let's try and solve it by first squaring both sides:
x² = 4.
The solutions are, clearly, x = 2 and x = -2. Yet, x = -2 is not a solution to the original equation x = 2, because -2 does not equal 2.

 

[zorro]

What part don't you understand? Do you not understand what the English sentence

Any solution to the original problem is also a solution to the new problem​

means? Do you not understand what I mean by "new problem" and "original problem"? Something else?

I did not understand what you meant by "new problem" and "original problem".

 

[tiny-tim]

Hi Abdul!

When we solve to equations say y² = 4x and x² + y² = 9/4 for x, we get two values -

x=0.5 and x=-4.5

The solutions to y² = 4x and x² + y² = 9/4

are (0.5, √2) (0.5, -√2), (-4.5, 3√2i), (-4.5, -3√2i).

If you're allowed imaginary numbers, then the last two solutions are ok.

If you're not allowed imaginary numbers, then the last two solutions are … not allowed!.

 

[Hurkyl]

I did not understand what you meant by "new problem" and "original problem".

Initially, you were trying to solve the problem

y^2 = 4x and x^2 + y^2 = 9/4​

By doing an algebraic manipulation (substituting 4x for y^2 in the second equation, I expect), you produced a new problem:

x^2 + 4x = 9/4​

You are guaranteed that if (x,y)=(a,b) for the first problem, then x=a is a solution for the second problem...

 

[zorro]

You are guaranteed that if (x,y)=(a,b) for the first problem, then x=a is a solution for the second problem...

So is x = -4.5 a solution? Do the curves intersect in the complex plane?

 

[tiny-tim]

sort-of …

if we let the y coordinate range over a "vertical" plane, perpendicular to the "horizontal" x-axis,

then y² = 4x is a horizontal parabola for x ≥ 0, touching a vertical parabola for x ≤ 0,

while x² + y² = 9/4 is a horizontal circle, touching a vertical hyperbola for |x| ≥ 3/2 …

the vertical parabola intersects the vertical hyperbola at (-4.5, 3√2i) and (-4.5, -3√2i)

 

[zorro]

Thank you tiny-tim.