+0  
 
+5
1073
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avatar+349 

Uhhh, finally. I tried posting this 32 times already. 32 TIMES!!!!. I kept accidentally quitting. SOO annoying. So anyway, hi.  Welcome to

                                                    DAILY QUESTION CONTEST #3 (very overdue)

 

 

(I like the new features,  it's been long since I've been)

 

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  \(x^2+y^2?\)

 

Question 2:

 

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

What is the value of  \(a^4+a^{-4}?\)

 

 

RULES/GUIDELINES:

 

1. A PERSON MAY ANSWER ONLY ONCE

2. FIRST 3 WILL BE RECOGNIZED AS WINNERS IN THE NEXT CONTEST

3. ANSWER REQUIRES SOLUTION

4. MOST NUMBER OF WINS AT THE END OF THE SEASON IS THE CHAMPION

 

GOOD LUCK! laugh

 Mar 8, 2016

Best Answer 

 #4
avatar+26364 
+5

Question 2:

 

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

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

 

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

 

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

 

laugh

 Mar 8, 2016
 #1
avatar+131 
0

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.

 Mar 8, 2016
 #2
avatar+131 
0

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. 

 Mar 8, 2016
 #8
avatar+349 
0

Well, this just tests the average person. There is no actual prize. The test is to see if a person has the average level of intelligence. smiley

Mathhemathh  Mar 8, 2016
 #3
avatar+26364 
+5

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  \(x^2+y^2?\)

 

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

 

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

 

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

 

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

 

laugh

 Mar 8, 2016
 #4
avatar+26364 
+5
Best Answer

Question 2:

 

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

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

 

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

 

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

 

laugh

heureka Mar 8, 2016
 #5
avatar+131 
0

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

 Mar 8, 2016
 #7
avatar+333 
0

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...

 Mar 8, 2016
edited by Misaki  Mar 8, 2016

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