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

 

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

supermanaccz  Sep 3, 2018
 #1
avatar+964 
+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: 

 

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

GYanggg  Sep 3, 2018
 #2
avatar+964 
+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. 

 

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

GYanggg  Sep 3, 2018

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