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# Help

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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

Sep 3, 2018

### 2+0 Answers

#1
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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:

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

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Sep 3, 2018
#2
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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.

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

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Sep 3, 2018