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\(\text{Given $x\neq0$, find the positive value of $b$ such that the equation $\frac 3x+\frac x3=b$ will have exactly one solution. }\)

 Aug 17, 2019
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\(\dfrac 3 x + \dfrac x 3 = b\\ \dfrac{9+x^2}{3x}=b\\ 9+x^2 = 3bx\\ x^2 - 3bx+9 = 0\\ \text{we could use the quadratic equation and do some further solving but we can see that if $b=2$}\\ \text{then we have $x^2-3bx+9 = (x-3)^2$, which provides exactly one solution, i.e. $x=3$}\)

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 Aug 17, 2019

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