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anemone
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Solve for the equation $a^3+3a^2-51a-107+54\sqrt{3}=0$
anemone said:Solve for the equation $a^3+3a^2-51a-107+54\sqrt{3}=0$
This took me ages to work out. Once I had solved the equation (by a very indirect method), I could see a much quicker way to verify the answer.anemone said:Solve for the equation $a^3+3a^2-51a-107+54\sqrt{3}=0$
Opalg said:This took me ages to work out. Once I had solved the equation (by a very indirect method), I could see a much quicker way to verify the answer.
[sp]Start with the fact (see here) that $\cos105^\circ = -\frac{\sqrt3-1}{2\sqrt2}$. Using the fact that $\cos(3\theta) = 4c^3 - 3c$, where $c = \cos\theta$, it follows that the solutions of the equation $4c^3 - 3c = -\frac{\sqrt3-1}{2\sqrt2}$ are $c = \cos35^\circ$, $c = \cos155^\circ$ and $c = \cos275^\circ$.
Now let $x = 6\sqrt2c$. The equation $4c^3 - 3c = -\frac{\sqrt3-1}{2\sqrt2}$ becomes $$ 4\Bigl(\frac x{6\sqrt2}\Bigr)^3 - 3\Bigl(\frac x{6\sqrt2}\Bigr) = -\frac{\sqrt3-1}{2\sqrt2}.$$ Multiply through by $54$ to get $x^3 - 54x = -54(\sqrt3-1).$
Next, let $a = x-1$. The equation becomes $(a+1)^3 - 54(a+1) = -54(\sqrt3-1),$ or $a^3 +3a^2 + 3a + 1 - 54a - 54 + 54\sqrt3 - 54 = 0.$ That simplifies to the given equation $a^3+3a^2-51a-107+54\sqrt{3}=0$. Since $a = x-1 = 6\sqrt2c-1$, the solutions of the equation are $$ a = 6\sqrt2\cos35^\circ - 1, \quad 6\sqrt2\cos155^\circ - 1, \quad 6\sqrt2\cos275^\circ - 1.$$
[You can check that these lie in the numerical intervals given by Albert.][/sp]
DreamWeaver said:Great thread, Anemone. As per usual... ;)(Ninja)(Sun)
DreamWeaver said:Keep 'em coming!(Hug)
anemone said:Thanks, DreamWeaver! And...welcome back to MHB!(Sun)(Coffee)
anemone said:I certainly will!(Star)
DreamWeaver said:Thoroughly glad to hear it. Personally, I think you add so much great material on here that we ought to rename it the "Anemone Board" (sorry, everyone else, but you know it makes sense ;) )
DreamWeaver said:...
Thoroughly glad to hear it. Personally, I think you add so much great material on here that we ought to rename it the "Anemone Board" (sorry, everyone else, but you know it makes sense ;) )
The first step in solving this equation is to isolate the radical term by subtracting 54√3 from both sides. This will leave you with a cubic equation of the form a^3+3a^2-51a-161=0. Then, using the Rational Root Theorem, you can determine possible rational roots and test them using synthetic division. Once you have found a rational root, you can factor the equation and use the quadratic formula to solve for the remaining roots.
Yes, it is possible to solve this equation using other methods such as factoring, completing the square, or using the cubic formula. However, the Rational Root Theorem is often the most efficient method for finding rational roots in a cubic equation.
Yes, this equation has a unique solution. Cubic equations can have up to three distinct solutions, but this particular equation only has one solution.
Yes, technology such as graphing calculators or computer programs can be used to solve this equation. They can help with finding rational roots, factoring, and solving for the remaining roots using the quadratic formula.
Yes, this equation has at least one real solution. This can be determined by graphing the equation and seeing where it intersects the x-axis.