+88 Find the product of all real values of \(r\) for which \(\dfrac{1}{2x}=\dfrac{r-x}{7}\) has exactly one real solution.
THANKS
+129849 Cross-multiply and we get that
2xr - 2x^2 = 7 rearrange as
2x^2 - 2rx + 7 = 0
Single real solutions will occur when the dicriminant = 0
So
(2r)^2 - ( 4) (7) (2) = 0
4r^2 - 56 = 0
4r^2 = 56 divide both sides by 4
r^2 = 14 take both roots
r = ±√14
And the product of these r's = (√14) ( - √14) = -14
