32 TIMES!!! (Daily Question Contest)

Question 2:

Suppose the number "a" satisfies the equation  \(a + \frac{1}{a} = 4\)

What is the value of  \(a^4 + \frac{1}{a^4}\) ?

\(\begin{array}{rcll} a+ \frac{1}{a} &=& 4 \qquad & | \qquad ()^2\\ \left( a+ \frac{1}{a} \right)^2 &=& 4^2 \\ a^2 + 2\cdot a\cdot \frac{1}{a} + \frac{1}{a^2} &=& 16 \\ a^2 + 2 + \frac{1}{a^2} &=& 16 \\ a^2 + \frac{1}{a^2} &=& 16-2 \\ \mathbf{ a^2 + \frac{1}{a^2} }& \mathbf{=} & \mathbf{14} \\ \end{array} \)

\(\begin{array}{rcll} a+ \frac{1}{a} &=& 4 \qquad & | \qquad ()^4\\ \left( a+ \frac{1}{a} \right)^4 &=& 4^4 \\ a^4 + 4 a^3 \cdot \frac{1}{a} + 6 a^2 \cdot \frac{1}{a^2} + 4 a \cdot \frac{1}{a^3} + \frac{1}{a^4} &=& 256 \\ a^4 + 4a^2 + 6 + \frac{4}{a^2} + \frac{1}{a^4} &=& 256 \\ a^4 + \frac{1}{a^4} + 4a^2 + 6 + \frac{4}{a^2} &=& 256 \\ a^4 + \frac{1}{a^4} + 4a^2 + \frac{4}{a^2} &=& 256 -6 \\ a^4 + \frac{1}{a^4} + 4a^2 +\frac{4}{a^2} &=& 250 \\ a^4 + \frac{1}{a^4} + 4 \left( a^2 + \frac{1}{a^2} \right) &=& 250 \qquad | \qquad a^2 + \frac{1}{a^2} = 14 \\ a^4 + \frac{1}{a^4} + 4 \cdot 14 &=& 250 \\ a^4 + \frac{1}{a^4} + 56 &=& 250 \\ a^4 + \frac{1}{a^4} &=& 250 -56\\ \mathbf{a^4 + \frac{1}{a^4} }&\mathbf{=}&\mathbf{ 194}\\ \end{array}\)

Q1) bit of a stupid question seeing as multiple answers are possible I will go with one for now as you dont exactly ask for more. Therefore if y=0 then x=4 so x^2+y^2 could just equal 16

Q2) a = 2+3^0.5 therefore the required value is 194.

I know I am a guest but I think your questions don't branch very far in terms of theoretical knowledge and are just simple algebra that most people with average intelligence could easly figure out. No one deserves a prize for answering these questions. 

Question 1:

If  \(y+4=(x-2)^2\)  and  \(x+4=(y-2)^2\)

with "x" not equal to "y",

what is the value of  \(x^2+y^2?\)

\(\begin{array}{lrcll} & y+4 &=& (x-2)^2 \\ & y +4 &=& x^2-4x+4\\ & y &=& x^2-4x\\ (1) & x^2 - 4x -y &=& 0 \\ \hline \\ & x+4 &=& (y-2)^2 \\ & x +4 &=& y^2-4y+4\\ & x &=& y^2-4y\\ (2) & y^2 - 4y -x &=& 0 \\ & y &=& \frac{4 \pm\sqrt{16-4(-x)}}{2} \\ & y &=& \frac{4 \pm\sqrt{16+4x}}{2} \\ & y &=& \frac{4 \pm \sqrt{4\cdot(4+x)}}{2} \\ & y &=& \frac{4 \pm 2\sqrt{4+x}}{2} \\ & y &=& 2 \pm \sqrt{4+x} \qquad \text{or} \qquad x = 2 \pm \sqrt{4+y}\\ \hline \\ & y &=& x^2-4x\\ & 2 \pm \sqrt{4+x}&=& x^2-4x\\ & \pm \sqrt{4+x}&=& x^2-4x-2 \qquad | \qquad \text{square both sides}\\ & 4+x&=& (x^2-4x-2)\cdot(x^2-4x-2)\\ & \dots \\ & x^4 -8x^3+12x^2+15x &=& 0 \\ \end{array}\)

\(\begin{array}{rcll} x(x^3 -8x^2+12x+15) &=& 0 \\ x_1 = 0 \\\\ x^3 -8x^2+12x+15 &=& 0 \\ x_2 = 5\\\\ (x-5)\cdot( x^2-3x-3) &=& 0 \\ x^2-3x-3 &=& 0 \\ x_3 &=& \frac{ 3+\sqrt{21} }{2} \qquad \text{ or } \qquad x_4 = \frac{ 3-\sqrt{21} }{2} \end{array}\)

\(\begin{array}{rcll} \text{ symmetry } x \text{ and } y :\\ y_1 &=& 0 \\ y_2 &=& 5 \\ y_3 &=& \frac{ 3+\sqrt{21} }{2} \\ y_4 &=& \frac{ 3-\sqrt{21} }{2} \end{array}\)

\(\begin{array}{rcll} x \ne y : \\ x_1 &=& \frac{ 3+\sqrt{21} }{2} \qquad y_1 &=& \frac{ 3-\sqrt{21} }{2} \\ x_2 &=& \frac{ 3-\sqrt{21} }{2} \qquad y_2 &=& \frac{ 3+\sqrt{21} }{2} \\\\ x^2+y^2 &=& 5\cdot(x+y) \\ &=& 5\cdot (x_1+y_1)\\ &=& 5\cdot (x_2+y_2) \\ &=& 15 \end{array}\)

What a tryhard response. Well done you graduated web2.0 idiot school

For question 1

Let \(y=x^2-4x\)

Let \(x=y^2-4y\)

With this \(x \neq y\)

We want to know what \(x^2+y^2\) so we square both of the equations to get that

\(x^2=x^4-8x^3+16x^2\) and with \(y \) we get

\(y^2=y^4-8y^3+16y^2\) Since there are differerent coefficents on both \(x\) and \(y\)

All we can do is put it in biggest degree to lowest degree

\(x^2+y^2=x^4+y^4−8x^3−8y^3+16x^2+16y^2\)

Question 2. \(4=a+\frac{1}{a}\) multiply both sides by a 

\(4a=a^2+1\) as you notice this a quadratic in disguise! hence

\(-a^2+4a-1=0\) as factorisation dosn't work we will use the quadratic formula

For values a=-1  b=4 c=-1 

\(a = {-4 \pm \sqrt{4^2-4} \over -2}\)simplify \(a = {2 \pm \sqrt{12} }\)

Simplify even more \(a=2 \pm 2\sqrt{3}\)

Now to find the value of \(a^4+a^{-4}\)\(a^4=97\pm56\sqrt{3}\)\(a^{-4}=\frac{7}{64}\pm \frac{1}{16}\sqrt{3}\)

\(a^4 + a^{-4}= 97 \pm 56\sqrt{3} + \frac{7}{64} \pm \frac{1}{16}\sqrt{3}\) after a bit more simplifying we get   

\(\frac{6215}{64}-\frac{897}{16}\sqrt{3}\) if \(97 - 56\sqrt{3} + \frac{7}{64} - \frac{1}{16}\sqrt{3}\) if \(97 + 56\sqrt{3} + \frac{7}{64} + \frac{1}{16}\sqrt{3}\) we get \(\frac{6215}{64}+\frac{897}{16}\sqrt{3}\) if \(97 + 56\sqrt{3} + \frac{7}{64} - \frac{1}{16}\sqrt{3}\) we get\(\frac{6215}{64}+\frac{895}{16}\sqrt{3}\) if

\(97 - 56\sqrt{3} + \frac{7}{64} + \frac{1}{16}\sqrt{3}\) we get \(\frac{6215}{64}-\frac{895}{16}\sqrt{3}\) that's all the combinations! PHEW...