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# math

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consider the intersection of the function; x2+y-8=0 and x+y=2. what is the greater of the x-coordinates of the points of intersection?

Guest May 15, 2017
#1
+20526
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consider the intersection of the function; x2+y-8=0 and x+y=2.

what is the greater of the x-coordinates of the points of intersection?

$$\begin{array}{|lrcll|} \hline (1) & x^2+y-8 &=& 0 \\ & y &=& 8-x^2 \\\\ (2) & x+y&=& 2 \quad & | \quad y = 8-x^2 \\ & x+8-x^2 &=& 2 \\ & x^2 - x - 6 &=& 0 \\ & x &=& \frac{1\pm \sqrt{1-4\cdot(-6)} }{2} \\ & x &=& \frac{1\pm \sqrt{1+24} }{2} \\ & x &=& \frac{1\pm 5 }{2} \\ \\ & x_1 &=& \frac{1+5 }{2} \\ & x_1 &=& \frac62 \\ &\mathbf{ x_1 } & \mathbf{=} & \mathbf{3} \\\\ & x_2 &=& \frac{1-5 }{2} \\ & x_2 &=& -\frac{4}{2}\\ &\mathbf{ x_2 } & \mathbf{=} & \mathbf{-2} \\ \hline \end{array}$$

The greater of the x-coordinates of the points of intersection is 3

heureka  May 15, 2017

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