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stuart clark

Modulus equation....

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in mathematics, a quadratic equation have two solutions when [latex]\Delta[/latex] is positive,

one solution if [latex]\Delta[/latex] is Zero, and No real solution if [latex]\Delta[/latex] is negative !

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It wouldn't lead to a proof, but if you're simply looking for an answer then plotting [imath]y(x)=||x^2-4x+3|-2|-m[/imath] for various positive values of [imath]m[/imath] will give it away. (you're looking for an answer of the form [imath]m>c[/imath] for some [imath]c[/imath]).

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Or just solve it as you would a normal equation with absolute values, but you'll end up having to solve four different quadratic equations.

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It's not difficult to plot the graph of [imath]y=||x^2-4x+3|-2|.[/imath] Start by plotting the graph of [imath]y=x^2-4x+3[/imath] and reflecting everything below the x-axis above it. Then drag it down vertically by two units and reflect everything below the x-axis above it again.

 

While graphs themselves do not constitute proofs, they do greatly help you get started. Doing the above, I make it that the graph above intersects the line [imath]y=m[/imath] at exactly 2 points precisely when [imath]m=0[/imath] or [imath]m>2.[/imath] It remains to justify this algebraically – which can be pretty tedious, but things do become simpler once you know what you need to do. ;)

Edited by shyvera

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