+654 If x^2 + y^2 +1/2 = x + y, find x^y+y^x
If |x|=3x+1, find (64x^2+48x+9)^2018
+981 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:
\(x^y+y^x\Rightarrow \frac12^\frac12+\frac12^\frac12\\ =\sqrt{\frac12}+\sqrt{\frac12}\\ =\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\\ =\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\boxed{\sqrt2}\)
+981 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.
\((8\cdot(-0.25)+3)^{4036}=(-2+3)^{4036}=1^{4036}=\boxed1\)
