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3 years ago

What are the solutions of -x^2+4x=x-4

Mathematics

1 answer:

Stolb23 [73]3 years ago

8 0

Answer:

x = -1 and x = 4

Step-by-step explanation:

Step 1: Get everything to one side of the equation so it's set equal to zero..

-x² + 4x = x - 4 becomes...

-x² + 3x + 4 = 0 (subtract x and add 4 to both sides)

x² - 3x - 4 = 0 (divide both sides by -1, we want the x² term to be positive)

(x - 4)(x + 1) = 0 (factor)

x - 4 = 0, then x = 4

and

x + 1 = 0, then x = -1

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Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

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Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

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Step-by-step explanation:

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Step-by-step explanation:

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