3 years ago

Locate√3 on number line and jusitify

Mathematics

1 answer:

alekssr [168]3 years ago

8 0

Root3 would be between 1 and 2 on the number line. This is known because 1 squared is 1, and 2 squared is 4.

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For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

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Simplify to,

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Then find solutions,

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$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

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the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

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Look at the drawing below.

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Answer:

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Answer:

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Step-by-step explanation:

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2 years ago

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Locate√3 on number line and jusitify​

Locate√3 on number line and jusitify