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Thanks!

- Thread starter drewfstr314
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- #1

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Thanks!

- #2

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The quartic formula! Didn't you learn it in school? :p

- #3

Hurkyl

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r is the root of [itex]x^4 + 48x^3 +536x^2 -956x + 372=0[/itex] between 0.6 and 0.7

Incidentally, I plugged your original equation and your quartic equation into wolframalpha and the solutions don't agree -- you've made an error in arithmetic or in transcription. (although correcting your arithmetic won't make things any easier)

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- #5

uart

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I'm pretty sure that the original equation has just one real solution. Write it as:_{1}≈0.996 and x_{2}≈-25.242, but the quartic has two extra solutions: at x_{3}≈0.607 and at x_{4}≈-24.361. These last two are extraneous solutions, but x_{1}and x_{2}are also solutions of the quartic.

[tex]\sqrt{7-x} = \frac{x^2}{2} - 12x + 10 [/tex]

and it's clear that the LHS is the upper half of a sideways parabola, with domain [itex]x \leq 7[/itex], while the RHS is a vertical parabola with axis of symmetry [itex]x=12[/itex]. Therefore, where the domains overlap both curves are monotonic. So there is only one real solution.

BTW. Fixed point iteration of [itex]x = \frac{1}{12} (0.5*x^2 + 10 - \sqrt{7-x} )[/itex] gives this solution at about 0.640259457069546.

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Your comment about one solution threw me off. We had gone over the problem (graphically, of course) in class, and there were two solutions. I then noticed that I had typed the wrong equation, and simply copied it from my first post. I switched the signs:I'm pretty sure that the original equation has just one real solution. Write it as:

[tex]\sqrt{7-x} = \frac{x^2}{2} - 12x + 10 [/tex]

and it's clear that the LHS is the upper half of a sideways parabola, with domain [itex]x \leq 7[/itex], while the RHS is a vertical parabola with axis of symmetry [itex]x=12[/itex]. Therefore, where the domains overlap both curves are monotonic.So there is only one real solution.

BTW. Fixed point iteration of [itex]x = \frac{1}{12} (0.5*x^2 + 10 - \sqrt{7-x} )[/itex] gives this solution at about 0.640259457069546.

[tex]\sqrt{7-x} = \frac{x^2}{2} + 12x - 10 [/tex]

should be correct. Wolfram Alpha confirms that there are two zeros here. Regardless of the signs, how would I find these solutions algebraically, without relying on the quartic formula?

Thanks!

Last edited:

- #7

uart

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Yes obviously that changes things. The axis of symmetry of that parabola is x = -12, so that it is now possible for the [itex]\sqrt{7-x}[/itex] curve to intersect with both "sides" of the parabola (I haven't checked that it actually does, but will take your word for it). With the previously given equation however, this was simply not possible.Your comment about one solution through me off. We had gone over the problem (graphically, of course) in class, and there were two solutions. I then noticed that I had typed the wrong equation, and simply copied it from my first post. I switched the signs:

[tex]\sqrt{7-x} = \frac{x^2}{2} + 12x - 10 [/tex]

should be correct. Wolfram Alpha confirms that there are two zeros here. Regardless of the signs, how would I find these solutions algebraically, without relying on the quartic formula?

Thanks!

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