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3 years ago

25) PLEASE HELP WITH QUESTION, WILL MARK BRAINLIEST + POINTS.

Mathematics

1 answer:

iogann1982 [59]3 years ago

4 0

You take the three zeros and put them back with the x s

(x-6)(x+5)(x-2) they are reversed when you put them back in

multiply the first two and get $x^2+5x -6x-30(x-2)$
simplify to $x^2-x-30(x-2)$

Multiply it out again and get $x^3-x^2-30x-2x^2+2x+60$
finally simplify and get $x^3-3x^2-28x+60$

So the answer is the first one

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$x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

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Similarly, because $(x - 4)$ and $(2\, x - 1)$ are two other inputs to the logarithm function in the original equation, they should also be positive. Therefore, $x > 4$ .

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$\displaystyle \ln (a) + \ln(b) = \ln\left(a \cdot b\right)$ .

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Left-hand side of this equation:

$\begin{aligned}&\ln(3\, x) - \ln(x - 4)= \ln\left(\frac{3\, x}{x -4}\right)\end{aligned}$

Right-hand side of this equation:

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$\displaystyle \frac{3\ x}{x - 4} = 3\, (2\, x - 1)$ .

Simplify and solve this equation for $x$ :

$x^2 - 5\, x + 2 = 0$ .

There are two real (but not rational) solutions to this quadratic equation: $\displaystyle \frac{5 + \sqrt{17}}{2}$ and $\displaystyle \frac{5 - \sqrt{17}}{2}$ .

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