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aev [14]
3 years ago
13

What is x+9=2(x_1)^2 in the form of ax^2+bx +c=0

Mathematics
1 answer:
Naddik [55]3 years ago
7 0
<span>x + 9 = 2(x - 1)^2
x + 9 = 2(x^2 - 2x + 1)
x + 9 = 2x^2 - 4x + 2
</span>2x^2 - 4x + 2 - x - 9 = 0
2x^2 - 5x -7 =0
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Answer:

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Step-by-step explanation:

See the sketch of the region in the attached graph.

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We solve it:

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\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx

Then using the basic rule to evaluate the integral:

\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4

Simplifying a bit:

\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4

Then plugging the limits of the integral:

\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]

Taking the root (rational exponents):

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Then doing those arithmetic computations we get:

\displaystyle\frac{37\pi}{10}

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