+0  
 
0
135
2
avatar+629 

If x^2 + y^2 +1/2 = x + y, find x^y+y^x

 

If |x|=3x+1, find (64x^2+48x+9)^2018

 Sep 3, 2018
 #1
avatar+971 
+2

If \(x^2 + y^2 +\frac12 = x + y\), find \(x^y+y^x\).

 

\(x^2 + y^2 +\frac12 = x + y\) Original, unsimplified equation
\( x^2-x+\frac14=-y^2+y-\frac14\) Moving terms to either side, \(\frac12\)split into \(\frac14's\)
\((x-\frac12)^2=-(y-\frac12)^2\) Rewrote the expressions as squares
\((x-\frac12)^2+(y-\frac12)^2=0\) Added \((y-\frac12)^2\) to both sides
\(x-\frac12=0\ \&\ y-\frac12=0\) Since \(n^2\ge0\), each expression is equal to 0
\(x=\frac12, y=\frac12\) Solved!

 

Now plugging our values into the expression: 

 

\(x^y+y^x\Rightarrow \frac12^\frac12+\frac12^\frac12\\ =\sqrt{\frac12}+\sqrt{\frac12}\\ =\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\\ =\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\boxed{\sqrt2}\)

.
 Sep 3, 2018
 #2
avatar+971 
+2

If \(|x|=3x+1\), find \((64x^2+48x+9)^{2018}\).

 

Rewriting expression:

 

\((64x^2+48x+9)^{2018},\\ =[(8x+3)^2]^{2018},\\ =(8x+3)^{4036}.\)

 

Solving for \(x\):

 

\(|x|=3x+1\\ x^2=(3x+1)^2\\ x^2=9x^2+6x+1\\ 8x^2+6x+1=0\\ (2x+1)(4x+1)=0\\ x_1=-0.5, x_2 = -0.25\)

 

After plugging our values into the equation to check if they work, we will see that only \(-0.25 \) satisfies the equation. 

 

\((8\cdot(-0.25)+3)^{4036}=(-2+3)^{4036}=1^{4036}=\boxed1\)

.
 Sep 3, 2018

10 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.