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3 years ago

Use the zero-product property to solve the following equation. -8(5n-2)=0

Mathematics

2 answers:

dem82 [27]3 years ago

8 0

Divide each side by -8:

$\frac{ - 8(5n - 2)}{ - 8} = \frac{0}{ - 8}$
$5n - 2 = 0$

Find n:
$5n = 2$
$n = \frac{2}{5}$

Aloiza [94]3 years ago

7 0

Answer:

The roots are {0, 2/5}.

Step-by-step explanation:

-8n(5n-2)=0 can be rewritten as (-8n)(5n-2) = 0.

This, in turn, can be rewritten as -8n = 0 (which yields the root n = 0) and 5n - 2 = 0 (which yields n = 2/5.

The roots are {0, 2/5}.

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Consider the differential equation

Ainat [17]

Answer:

$W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )=e^x$

Step-by-step explanation:

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$4y''-4y'+y=0\\4\left (\frac{1}{4}e^{\frac{x}{2}} \right )-4\left ( \frac{1}{2}e^{\frac{x}{2}}\right )+e^{\frac{x}{2}}\\=e^{\frac{x}{2}}-2e^{\frac{x}{2}}+e^{\frac{x}{2}}\\=2e^{\frac{x}{2}}-2e^{\frac{x}{2}}\\=0$

So, $y=e^{\frac{x}{2}}$ is the solution of the given equation.

Now, let $y=xe^{\frac{x}{2}}$

Differentiate with respect to $x$

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

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Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}} \right )-4\left ( e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )\right )+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-2e^{\frac{x}{2}}(2+x)+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-4e^{\frac{x}{2}}-2xe^{\frac{x}{2}}+xe^{\frac{x}{2}}\\=0$

To find: $W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )$

Solution:

Let $u=e^{\frac{x}{2}}\,,\,v=xe^{\frac{x}{2}}$

$W(u,v)=\left | \begin{matrix}u&v\\u'&v' \end{matrix} \right |\\=\left | \begin{matrix}e^{\frac{x}{2}}&xe^{\frac{x}{2}}\\\frac{1}{2}e^{\frac{x}{2}}&e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \end{matrix} \right |\\=e^{\frac{x}{2}}\left [ e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \right ]-\frac{1}{2}e^{\frac{x}{2}}xe^{\frac{x}{2}}\\=e^x\left ( 1+\frac{x}{2} \right )-\frac{1}{2}xe^x\\=e^x+\frac{1}{2}xe^x-\frac{1}{2}xe^x\\=e^x$

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