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\(Find the product of all real values of r for which \frac{1}{2x}=\frac{r-x}{7} has exactly one real solution.\)

 Apr 24, 2019
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\(\dfrac{1}{2x} = \dfrac{r-x}{7}\\ 7 = 2rx - 2x^2\\ 2x^2 - 2rx + 7 = 0\\ \text{There will be exactly one real solution if the discriminant is equal to 0}\\ D = (-2r)^2 - 4(2)(7) = 0\\ 4r^2 = 56\\ r^2 = 14\\ r = \pm \sqrt{14}\\ \sqrt{14}(-\sqrt{14}) = -14\)

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 Apr 24, 2019

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