1. ## Challenge Algebra Equations

$\displaystyle \text{Solve for \emph{x}. Real solutions only.}$

$\displaystyle 64x^6-112x^4+56x^2-7=2\sqrt{1-x^2}$

$\displaystyle x^5-15x^3-45x-27=0$

2. ## Re: Challenge Algebra Equations

Wait, are the two equations separate questions or does x have to satisfy both equations?

3. ## Re: Challenge Algebra Equations

Originally Posted by Mrdavid445
Wait, are the two equations separate questions or does x have to satisfy both equations?
They are two separate equations. Solve for x in each equation. Have fun!

4. ## Re: Challenge Algebra Equations

Originally Posted by thevinh
$\displaystyle \text{Solve for \emph{x}. Real solutions only.}$
$\displaystyle x^5-15x^3-45x-27=0$
x = 4.22786... or -4.15136... or -.54664... ; other 2 are imaginary.

Change the -27 to +116, then x = 4 exactly.

Any purpose to this madness ?

5. ## Re: Challenge Algebra Equations

Wolfram Alpha ftw.

6. ## Re: Challenge Algebra Equations

Originally Posted by Wilmer
x = 4.22786... or -4.15136... or -.54664... ; other 2 are imaginary.

Change the -27 to +116, then x = 4 exactly.

Any purpose to this madness ?
$\displaystyle \text{Hi, I realized I made a mistake in posting the equation. I am so sorry about that. The solution is posted below.}$

$\displaystyle \textbf{Problem 2:}\quad \text{Solve}\mspace{7mu} x^5-15x^3+45x-27=0 \quad\quad (1)$

$\displaystyle \text{From the Rational Root Theorem,}\mspace{5mu} x=-3,\mspace{5mu} \text{is one of the many solution}.$

$\displaystyle \text{Thus}\mspace{7mu} (1)\, \Leftrightarrow\, (x+3)(x^4-3x^3-6x^2+18x-9)=0$

$\displaystyle x^4-3x^3-6x^2+18x-9=0\, \Leftrightarrow\, x^4-3x^2(x-1)-9(x-1)^2=0$

$\displaystyle \text{Let}\mspace{7mu} y=x-1\, \Rightarrow\, x^4-3yx^2-9y^2=0\quad \quad (2)$

$\displaystyle \text{Solve for \emph{x} in terms of \emph{y}, we get:}\mspace{7mu} 2x^2=3y\pm3y\sqrt{5}$

$\displaystyle \text{\underline{Case 1.}}\quad 2x^2=3y+3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)+3\sqrt{5}(x-1)$

$\displaystyle \Leftrightarrow\, 2x^2-3\left(\sqrt{5}+1\right)x+3\left(\sqrt{5}+1\right) =0$

$\displaystyle \text{Solve the quadratic equation, we get:}$

$\displaystyle x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)$

$\displaystyle \text{\underline{Case 2.}}\quad 2x^2=3y-3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)-3\sqrt{5}(x-1)$

$\displaystyle \Leftrightarrow\, 2x^2+3\left(\sqrt{5}-1\right)x-3\left(\sqrt{5}-1\right)=0$

$\displaystyle \text{Solve the quadratic equation, we get:}$

$\displaystyle x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)$

$\displaystyle \text{So (1) has five real solutions:}$

$\displaystyle x=-3$

$\displaystyle x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)$

$\displaystyle x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)$

7. ## Re: Challenge Algebra Equations

Originally Posted by thevinh
$\displaystyle \text{Hi, I realized I made a mistake in posting the equation. I am so sorry about that. The solution is posted below.}$

$\displaystyle \textbf{Problem 2:}\quad \text{Solve}\mspace{7mu} x^5-15x^3+45x-27=0 \quad\quad (1)$

$\displaystyle \text{From the Rational Root Theorem,}\mspace{5mu} x=-3,\mspace{5mu} \text{is one of the many solution}.$

$\displaystyle \text{Thus}\mspace{7mu} (1)\, \Leftrightarrow\, (x+3)(x^4-3x^3-6x^2+18x-9)=0$

$\displaystyle x^4-3x^3-6x^2+18x-9=0\, \Leftrightarrow\, x^4-3x^2(x-1)-9(x-1)^2=0$

$\displaystyle \text{Let}\mspace{7mu} y=x-1\, \Rightarrow\, x^4-3yx^2-9y^2=0\quad \quad (2)$

$\displaystyle \text{Solve for \emph{x} in terms of \emph{y}, we get:}\mspace{7mu} 2x^2=3y\pm3y\sqrt{5}$

$\displaystyle \text{\underline{Case 1.}}\quad 2x^2=3y+3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)+3\sqrt{5}(x-1)$

$\displaystyle \Leftrightarrow\, 2x^2-3\left(\sqrt{5}+1\right)x+3\left(\sqrt{5}+1\right) =0$

$\displaystyle \text{Solve the quadratic equation, we get:}$

$\displaystyle x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)$

$\displaystyle \text{\underline{Case 2.}}\quad 2x^2=3y-3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)-3\sqrt{5}(x-1)$

$\displaystyle \Leftrightarrow\, 2x^2+3\left(\sqrt{5}-1\right)x-3\left(\sqrt{5}-1\right)=0$

$\displaystyle \text{Solve the quadratic equation, we get:}$

$\displaystyle x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)$

$\displaystyle \text{So (1) has five real solutions:}$

$\displaystyle x=-3$

$\displaystyle x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)$

$\displaystyle x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)$

It's looking right and but to be sure we have to wait thread starter to see if he has another method to solve this.algebra word problem solver free