MHB Solve Equation: $a^3+3a^2-51a-107+54\sqrt{3}=0$

[anemone]

Solve for the equation $a^3+3a^2-51a-107+54\sqrt{3}=0$

 

[Albert1]

Solve for the equation $a^3+3a^2-51a-107+54\sqrt{3}=0$

Spoiler

using Descartes' rule of signs ,and intermediate value theorem we know this equation must have 1 positive root
and two negative roots
f(5)f(6)<0 (5<a positive root<6 )
f(0)f(-1)<0 (-1<a negative root<0)
f(-8)(-9)<0(-9<another negative root<-8)
we can use Newton's method to find more accurate roots of the equation

 

[Opalg]

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.

[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]

 

[Saitama]

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]

This is just awesome! (Clapping) (Bow)

 

[anemone]

Thanks, Albert for your suggestion of the use of numerical method to tackle this problem.:)

Hey Opalg, I really like the way how you verified the answer, well done Opalg and thanks for participating to both of you!

Solution suggested by other:

Spoiler

First, we manipulate the original equation $a^3+3a^2-51a-107+54\sqrt{3}=0$ by rewriting it as below:

$a^3+3a^2+(3a+1)-(3a+1)-51a-107+54\sqrt{3}=0$

$(a^3+3a^2+3a+1)-54a-108+54\sqrt{3}=0$

$(a+1)^3-54a-54-54+54\sqrt{3}=0$

$(a+1)^3-54(a+1)-54(1-\sqrt{3})=0$

$b^3-54b-54(1-\sqrt{3})=0$(*) if $b=a+1$

If we want to further solve the equation, we have to resort to use the triple angle formula for cosine function since $\cos 3x=4\cos^3x-3\cos x$, i.e., we must make $b^3-54b$ to take the form of $4\cos^3x-3\cos x$ and this could be done by observing that:

$\dfrac{b^3}{54b}=\dfrac{4\cos^3 x}{3\cos x}$ and this gives $b=\sqrt{72} \cos x$ and substituting it back to (*) we get:

$(\sqrt{72} \cos x)^3-54(\sqrt{72} \cos x)-54(1-\sqrt{3})=0$

$72\sqrt{72} \cos^3 x-54\sqrt{72} \cos x-54(1-\sqrt{3})=0$

$18\sqrt{72}(4 \cos^3 x-3\cos x)-54(1-\sqrt{3})=0$

$18\sqrt{72}\cos 3x-54(1-\sqrt{3})=0$

$\begin{align*}\cos 3x&=\dfrac{54(1-\sqrt{3})}{18\sqrt{72}}\\&=\dfrac{-(\sqrt{3}-1)}{2\sqrt{2}}\end{align*}$

Remember that

$\begin{align*}\cos 75^{\circ}&=\sqrt{\dfrac{\cos 150^{\circ}+1}{2}}\\&=\sqrt{\dfrac{-\dfrac{\sqrt{3}}{2}+1}{2}}\\&=\dfrac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}\\&=\dfrac{\sqrt{3-2\sqrt{3}+1}}{2\sqrt{2}}\\&=\dfrac{\sqrt{(\sqrt{3}-1)^2}}{2\sqrt{2}}\\&=\dfrac{\sqrt{3}-1}{2\sqrt{2}} \end{align*}$

So, $\cos (180^{\circ}-75^{\circ})=\cos 105^{\circ}=\dfrac{-(\sqrt{3}-1)}{2\sqrt{2}}$ and the other two smallest positive values for $3x$ are $360^{\circ}-105^{\circ}=255^{\circ}$ and $360^{\circ}+105^{\circ}=465^{\circ}$ and this gives $x=35^{\circ},\,85^{\circ},\,155^{\circ},\,$ or $a=\sqrt{72}\cos 35^{\circ}-1,\,\sqrt{72}\cos 85^{\circ}-1,\,\sqrt{72}\cos 155^{\circ}-1$.

 

[DreamWeaver]

Great thread, Anemone. As per usual... ;)(Ninja)(Sun)

Keep 'em coming!(Hug)

 

[anemone]

Great thread, Anemone. As per usual... ;)(Ninja)(Sun)

Thanks, DreamWeaver! And...welcome back to MHB!(Sun)(Coffee)

Keep 'em coming!(Hug)

I certainly will!(Star)

 

[DreamWeaver]

Thanks, DreamWeaver! And...welcome back to MHB!(Sun)(Coffee)

Thank you...! (Sun) It's great to be back!

I certainly will!(Star)

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 ;) )

 

[anemone]

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 ;) )

I'm enormously flattered that you like my challenge posts...but, there are also those who took part in my challenges that ought to gain some sort of recognition as well!:p Also, this site main function is for the students from the four corners of the world to asking for help when they encountered difficulty in solving any of the mathematics problems, my posts are really only serve to complement MHB...:D But I truly know you mean well...(Star) Thanks, buddy!(Sun)

Overall, MHB is a very good name and actually the third letter $B$ means very much to me because that is also the first letter of my nick name!(Sun)

 

[MarkFL]

...
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 ;) )

anemone's challenge problems are certainly a staple of MHB. We are very pleased that she works so hard to do this for MHB. (Clapping)(Sun)