- #1
nobahar
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Homework Statement
I need to find the range for y in the quadratic [tex]x^2+2x+3=y[/tex]
Using the determinant [tex]b^2-4ac [/tex] where [tex]ax^2+bx+c[/tex]
Homework Equations
[tex]x^2+2x+3=y[/tex]
The Attempt at a Solution
Okay, so:
To use the determinant [tex] y=0, x^2+2x+3=0[/tex], in which case [tex] b^2-4.a.c = -8 [/tex]
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,
[tex]x^2+2x+3=y[/tex] gives [tex]x^2+2x+3-y=0[/tex], in which case the determinant, [tex]2^2-4.1.(3-y)=4-12+4y[/tex]
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.