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3 years ago

How do I solve 2x^2-32=0 using the square root method?

Mathematics

2 answers:

Ainat [17]3 years ago

3 0

Answer:

x = -4,4

Step-by-step explanation:

2x^2 - 32 = 0

2x^2 = 32

x^2 = 16

x = either -4, 4

If there is only space for one answer put 4.

Hope this helped!

raketka [301]3 years ago

3 0

Answer: 4 or -4

Step-by-step explanation:

2x^2-32+32=0+32

2x^2=32

1/2*2x^2=32*1/2

x^2=16

x=4 or -4

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Plz use Differential equation method to solve this:

Oduvanchick [21]

Answer:

Step-by-step explanation:

We have the differential equation $y' = y + \frac{x}{y}$ with initial conditions $y(0)=1$ .

First, notice that the equation can be rewritten as

$y'-y =xy^{-1}$ ,

which is a Bernoulli equation. Once we have recognized the type of the equation we know how to continue. Recall that a Bernoulli equation has the general form

$y'+p(x)y=q(x)y^n$ .

In this particular case we have $n=-1$ . This kind of equation is solved by the change of variable $z=y^{1-n}$ . In our exercise we get $z=y^{1-(-1)}=y^2$ . Now we take derivatives and get

$z'=2yy'$ which es equivalent to $\frac{z'}{2y}=y'$ .

Then, we substitute the value of $y'$ we have obtained in the original equation:

$\frac{z'}{2y}-y = xy^{-1}$ .

The next step is to multiply the whole equation by $2y$ , in order to eliminate the denominator of $z'$ . Thus,

$z' -2y^2=2x$ .

Recall that $y^2=z$ , then

$z' -2z=2x$ .

This last equation is a linear equation, which has general solution

$z(x) = \exp\left(-\int(-2)dx\right)\left(\int 2x \exp\left(\int(-2)dx + C\right)\right)$ .

So, let us calculate the integral that appear in the formula:

$\int(-2)dx = -2x$

$\int 2x e^{-2x}dx = -\frac{\left(2x+1\right)e^{-2x}}{2}$ .

Then, the solution for $z$ is

$z(x) = e^{2x}\left(-\frac{\left(2x+1\right)e^{-2x}}{2} + C\right) = -\frac{\left(2x+1\right)}{2} + Ce^{2x}$ .

Now, we return the change of variable:

$(y(x))^2 =-\frac{\left(2x+1\right)}{2} + Ce^{2x}$ .

The last step is to find the value of the constant $C$ . In order to do this, substitute the initial value:

$(y(0))^2 = 1 =\frac{\left(2\cdot 0+1\right)}{2} + Ce^{2\cdot 0} = -\frac{1}{2} + C$ .

Thus, we have the equation

$1=-\frac{1}{2} + C$ that gives $C=\frac{3}{2}$ .

Therefore,

$(y(x))^2 = -\frac{\left(2x+1\right)}{2} + \frac{3}{2}e^{2x}$ .

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Answer:

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Step-by-step explanation:

Hey there!

I am going to use this image as an example so you can understand the formula

The formula for problems like these is

B(BA)=D(DC)

So all we have to do is plug in the numeric values

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now we just solve for x

5(5+x+3)=6(6+x)

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so now we have

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