Determinants in quadratics and the range of y

[nobahar]

Homework Statement

I need to find the range for y in the quadratic $$x^2+2x+3=y$$
Using the determinant $$b^2-4ac$$ where $$ax^2+bx+c$$

Homework Equations

$$x^2+2x+3=y$$

The Attempt at a Solution

Okay, so:
To use the determinant $$y=0, x^2+2x+3=0$$, in which case $$b^2-4.a.c = -8$$
So there are no real roots (I'm still unsure whether this is of any use anyway, unless the determinant =0, in which case I would have the maximum or minimum for the function and therefore the range would be from this point onwards, i.e. y>=0 or y<=0). But since there are no real roots...
However,
$$x^2+2x+3=y$$ gives $$x^2+2x+3-y=0$$, in which case the determinant, $$2^2-4.1.(3-y)=4-12+4y$$
Now the book suggests that this determinant should be an inequality, namely >=0, but surely, as I have subtracted y, the determinant should equal zero: -8+4y=0?
I don't think I've understood what's going on, but hopefully my attempt highlights my misundertstanding. I can solve the problem following the guideline, but I want to know WHY it is the way the book says. It seems, by subtracting y, Iam in essence 'moving the graph down' (?), although all values of course yield 0, as y is a variable (and the equation says so!). But by subtracting y and using the determinant, I am somehow identifying the lowest value for y, and shouldn't the determinant =0; after all, there can't be more than one root...
I hope this is clear, I really need to get this, so I can elaborate or alter the wording or something if its not clear...
Thanks everyone.

 

[diazona]

It's actually called the discriminant (determinant is a matrix thing). But anyway: by subtracting y you are indeed moving the graph down, and the point of that is to move it down far enough that it intersects the horizontal axis. If y takes the precise value that moves the curve down just far enough to touch the x axis, then yes, the discriminant will be zero. But you're forgetting about the possibility of y taking a larger value, and in that case there would be two roots since you've moved the graph so far down that its minimum actually lies below the axis. The discriminant would then be positive. That's why the book tells you to use an inequality.

Hopefully that helps a little - I know it's not the clearest explanation.

 

[nobahar]

ohhh, so I could keep making y larger in the DISCRIMINANT (oops!), $$-8+4y$$ until it hits the x-axis, at disc=o, which gives one root, for the minimun value of y, and then passes through the x-axis, disc>0, which gives all subsequent values of y, with two roots, and hence the range of y?
Thanks, it was nice and clear, and thanks for the speedy response!

 

[HallsofIvy]

Homework Statement

I need to find the range for y in the quadratic $$x^2+2x+3=y$$
Using the determinant $$b^2-4ac$$ where $$ax^2+bx+c$$

Homework Equations

$$x^2+2x+3=y$$

The Attempt at a Solution

Okay, so:
To use the determinant $$y=0, x^2+2x+3=0$$, in which case $$b^2-4.a.c = -8$$
So there are no real roots (I'm still unsure whether this is of any use anyway, unless the determinant =0, in which case I would have the maximum or minimum for the function and therefore the range would be from this point onwards, i.e. y>=0 or y<=0). But since there are no real roots...
However,
$$x^2+2x+3=y$$ gives $$x^2+2x+3-y=0$$, in which case the determinant, $$2^2-4.1.(3-y)=4-12+4y$$
Now the book suggests that this determinant should be an inequality, namely >=0, but surely, as I have subtracted y, the determinant should equal zero: -8+4y=0?
I don't think I've understood what's going on, but hopefully my attempt highlights my misundertstanding. I can solve the problem following the guideline, but I want to know WHY it is the way the book says. It seems, by subtracting y, Iam in essence 'moving the graph down' (?), although all values of course yield 0, as y is a variable (and the equation says so!). But by subtracting y and using the determinant, I am somehow identifying the lowest value for y, and shouldn't the determinant =0; after all, there can't be more than one root...
I hope this is clear, I really need to get this, so I can elaborate or alter the wording or something if its not clear...
Thanks everyone.

Yep, "discriminant", not "determinant". There will be two real roots if the discriminant is positive, none if the discrimant is negative and exactly one (a double root) if the discrimant is zero. Now, think about what the graph of $y= x^2$ looks like! It is a parabola going down to a vertex, then back up again. Drawing horizontal lines you should see that the parabola crosses the line twice (two real roots) if the line is above the vertex, does not cross the line (no real roots) if the line is above the vertex, and exactly one root if the horizontal line is at the vertex.

That is, the vertex occurs at that value of y that gives one root: that makes the discriminant 0, exactly as you say. So: what value of y makes 4- 12+ 4y= 0?

 

[VietDao29]

Now the book suggests that this determinant should be an inequality, namely >=0, but surely, as I have subtracted y, the determinant should equal zero: -8+4y=0?

No, your reasoning does not sound good, subtracting y does not means that the discriminant should be 0.

Ok, so the book ask you to find the range of y, in x² + 2x + 3 = y. Right?

So, say y₀ is in that range, that means, there exists some x₀, such that: x₀² + 2x₀ + 3 = y₀.

Or in another word, the equation: x² + 2x + 3 - y₀ = 0 must have at least one solution (one of the solution(s) must be x₀)

So, to find the range of y in that problem, we have to find all y, which makes the equation: x² + 2x + 3 - y = 0 have at least one solution, which means, the discriminant of this equation must be equal or greater than 0.

Hope I'm being clear enough. :)

If you find anywhere unclear, don't hesitate to ask. :)

 

[nobahar]

Thanks VietDao. I think I got it: and if this is so, then my second post is consistent with your response.
HallsofIvy: the only value y can take is 2. Forgive me if this is not the question!
Thanks for the responses.