Solve this equation using the radical equation method

Mathematics

1 answer:

Yuki888 [10]3 years ago

6 0

The domain:
$3x+6\geq0\ \ |-6\\3x\geq-6\ \ \ |:3\\x\geq-2\\\\D:x\in\left< -2;\ \infty\right)$

$2x=\sqrt{3x+6}-1\ \ \ |+1\\\\2x+1=\sqrt{3x+6}\ \ \ \ |^2\\\\(2x+1)^2=\left(\sqrt{3x+6}\right)^2\ \ \ |use:\ (a+b)^2=a^2+2ab+b^2\\\\(2x)^2+2\cdot2x\cdot1+1^2=3x+6\\\\4x^2+4x+1=3x+6\ \ \ \ |-3x-6\\\\4x^2+x-5=0\\\\4x^2-4x+5x-5=0\\\\4x(x-1)+5(x-1)=0\\\\(x-1)(4x+5)=0\iff x-1=0\ \vee\ 4x+5=0\\\\x=1\ \vee\ 4x=-5\\\\x=1\in D\ \vee\ x=-1.25\in D$

$Answer:\ \boxed{x=-1.25\ \vee\ x=-1}$

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Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

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Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

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