9514 1404 393
Answer:
2x² +5x -12 = 0
Step-by-step explanation:
When p and q are roots of a quadratic, its factored form can be written as ...
(x -p)(x -q) = 0
Here, the roots are given as -4 and 3/2, so the factored form would be ...
(x -(-4))(x -(3/2) = 0
Multiplying by 2 gives us ...
(x +4)(2x -3) = 0
Expanding the product, we find the desired quadratic is ...
2x² +5x -12 = 0
Coordinates of the midpoint of AC:
M ( (-6-2) / 2) , ( 7-9 ) /2 ) = ( -4, -1 )
d ( BM ) = √ ( 4 + 4 )² + ( -1+ 1 )²
d ( BM ) = √ 8 ² =√ 64 = 8
The length of the median from angle B is 8.
-30x cubed
There are three x hence the cubes and then all you do is multiply (5)(-2)and(3) together to get -30
First substitute
to rewrite the integral as
![\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = \int_0^1 \frac{\sqrt{y-y^2}}{1+y^2} \, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cpi%2F4%7D%20%5Csqrt%7B%5Ctan%28x%29%20-%20%5Ctan%5E2%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint_0%5E1%20%5Cfrac%7B%5Csqrt%7By-y%5E2%7D%7D%7B1%2By%5E2%7D%20%5C%2C%20dy)
Now use an Euler substitution,
to rewrite it again as
![\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = 2 \int_0^\infty \frac{t^2}{(t^2+1)^2 + 1) (t^2 + 1)} \, dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cpi%2F4%7D%20%5Csqrt%7B%5Ctan%28x%29%20-%20%5Ctan%5E2%28x%29%7D%20%5C%2C%20dx%20%3D%202%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bt%5E2%7D%7B%28t%5E2%2B1%29%5E2%20%2B%201%29%20%28t%5E2%20%2B%201%29%7D%20%5C%2C%20dt)
where we take
![\sqrt{y - y^2} = \sqrt{-y(y-1)} = yt \implies y = \dfrac1{1+t^2} \text{ and } dy = -\dfrac{2t}{(1+t^2)^2} \, dt](https://tex.z-dn.net/?f=%5Csqrt%7By%20-%20y%5E2%7D%20%3D%20%5Csqrt%7B-y%28y-1%29%7D%20%3D%20yt%20%5Cimplies%20y%20%3D%20%5Cdfrac1%7B1%2Bt%5E2%7D%20%5Ctext%7B%20and%20%7D%20dy%20%3D%20-%5Cdfrac%7B2t%7D%7B%281%2Bt%5E2%29%5E2%7D%20%5C%2C%20dt)
Partial fractions:
![\displaystyle \frac{t^2}{((t^2+1)^2+1) (t^2 + 1)} = \dfrac{t^2+2}{t^4+2t^2+2} - \dfrac1{t^2+1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bt%5E2%7D%7B%28%28t%5E2%2B1%29%5E2%2B1%29%20%28t%5E2%20%2B%201%29%7D%20%3D%20%5Cdfrac%7Bt%5E2%2B2%7D%7Bt%5E4%2B2t%5E2%2B2%7D%20-%20%5Cdfrac1%7Bt%5E2%2B1%7D)
so that
![\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = 2 \left(\int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt - \int_0^\infty \frac{dt}{t^2+1}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cpi%2F4%7D%20%5Csqrt%7B%5Ctan%28x%29%20-%20%5Ctan%5E2%28x%29%7D%20%5C%2C%20dx%20%3D%202%20%5Cleft%28%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bt%5E2%2B2%7D%7Bt%5E4%2B2t%5E2%2B2%7D%20%5C%2C%20dt%20-%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bdt%7D%7Bt%5E2%2B1%7D%5Cright%29)
The second integral is trivial,
![\displaystyle \int_0^\infty \frac{dt}{t^2+1} = \lim_{t\to\infty}\tan^{-1}(t) - \tan^{-1}(0) = \frac\pi2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bdt%7D%7Bt%5E2%2B1%7D%20%3D%20%5Clim_%7Bt%5Cto%5Cinfty%7D%5Ctan%5E%7B-1%7D%28t%29%20-%20%5Ctan%5E%7B-1%7D%280%29%20%3D%20%5Cfrac%5Cpi2)
For the other, I'm compelled to use the residue theorem, though real methods are doable too (e.g. trig substitution). Consider the contour integral
![\displaystyle \int_\Gamma f(z) \, dz = \int_\Gamma \frac{z^2+2}{z^4+2z^2+2} \, dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%5CGamma%20f%28z%29%20%5C%2C%20dz%20%3D%20%5Cint_%5CGamma%20%5Cfrac%7Bz%5E2%2B2%7D%7Bz%5E4%2B2z%5E2%2B2%7D%20%5C%2C%20dz)
where
is a semicircle in the upper half of the complex plane, and its diameter lies on the real axis connecting
to
. The value of this integral is 2πi times the sum of the residues in the upper half-plane. It's fairly straightforward to convince ourselves that the integral along the circular arc vanishes as
, so the contour integral converges to the integral over the entire real line. Note that
![\displaystyle 2 \int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt = \int_{-\infty}^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bt%5E2%2B2%7D%7Bt%5E4%2B2t%5E2%2B2%7D%20%5C%2C%20dt%20%3D%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20%5Cfrac%7Bt%5E2%2B2%7D%7Bt%5E4%2B2t%5E2%2B2%7D%20%5C%2C%20dt)
since the integrand is even.
Find the poles of
.
![z^4 + 2z^2 + 2 = 0 \\\\ ~~~~ \implies (z^2+1)^2 = -1 \\\\ ~~~~ \implies z^2 = -1 \pm i \\\\ ~~~~ \implies z = \pm \sqrt{-1 \pm i} = \sqrt[4]{2}\, e^{\pm i(3\pi/8 + \pi k)}](https://tex.z-dn.net/?f=z%5E4%20%2B%202z%5E2%20%2B%202%20%3D%200%20%5C%5C%5C%5C%20~~~~%20%5Cimplies%20%28z%5E2%2B1%29%5E2%20%3D%20-1%20%5C%5C%5C%5C%20~~~~%20%5Cimplies%20z%5E2%20%3D%20-1%20%5Cpm%20i%20%5C%5C%5C%5C%20~~~~%20%5Cimplies%20z%20%3D%20%5Cpm%20%5Csqrt%7B-1%20%5Cpm%20i%7D%20%3D%20%5Csqrt%5B4%5D%7B2%7D%5C%2C%20e%5E%7B%5Cpm%20i%283%5Cpi%2F8%20%2B%20%5Cpi%20k%29%7D)
where
.
The two poles we care about are at
and
. Compute the residues at each one.
![\displaystyle \mathrm{Res}\left\{f(z),z=z_1\right\} = \lim_{z\to z_1} \frac{f(z)}{z-z_1} = -\frac1{2^{7/4}}\,ie^{-i\,\pi/8} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~= -\frac1{2^{7/4}} \left(\sin\left(\frac\pi8\right) + i \cos\left(\frac\pi8\right)\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmathrm%7BRes%7D%5Cleft%5C%7Bf%28z%29%2Cz%3Dz_1%5Cright%5C%7D%20%3D%20%5Clim_%7Bz%5Cto%20z_1%7D%20%5Cfrac%7Bf%28z%29%7D%7Bz-z_1%7D%20%3D%20-%5Cfrac1%7B2%5E%7B7%2F4%7D%7D%5C%2Cie%5E%7B-i%5C%2C%5Cpi%2F8%7D%20%5C%5C%5C%5C%20~~~~~~~~~~~~~~~~~~~~~~~~%3D%20-%5Cfrac1%7B2%5E%7B7%2F4%7D%7D%20%5Cleft%28%5Csin%5Cleft%28%5Cfrac%5Cpi8%5Cright%29%20%2B%20i%20%5Ccos%5Cleft%28%5Cfrac%5Cpi8%5Cright%29%5Cright%29)
![\displaystyle \mathrm{Res}\left\{f(z),z=z_2\right\} = \lim_{z\to z_2} \frac{f(z)}{z-z_2} = -\frac1{2^{7/4}}\,ie^{i\,\pi/8} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~= \frac1{2^{7/4}} \left(\sin\left(\frac\pi8\right) - i \cos\left(\frac\pi8\right)\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmathrm%7BRes%7D%5Cleft%5C%7Bf%28z%29%2Cz%3Dz_2%5Cright%5C%7D%20%3D%20%5Clim_%7Bz%5Cto%20z_2%7D%20%5Cfrac%7Bf%28z%29%7D%7Bz-z_2%7D%20%3D%20-%5Cfrac1%7B2%5E%7B7%2F4%7D%7D%5C%2Cie%5E%7Bi%5C%2C%5Cpi%2F8%7D%20%5C%5C%5C%5C%20~~~~~~~~~~~~~~~~~~~~~~~~%3D%20%5Cfrac1%7B2%5E%7B7%2F4%7D%7D%20%5Cleft%28%5Csin%5Cleft%28%5Cfrac%5Cpi8%5Cright%29%20-%20i%20%5Ccos%5Cleft%28%5Cfrac%5Cpi8%5Cright%29%5Cright%29)
By the residue theorem,
![\displaystyle \int f(z) \, dz = 2\pi i \sum_{\rm poles} \mathrm{Res}\{f(z)\} = \frac{4\pi}{2^{7/4}} \cos\left(\frac\pi8\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20f%28z%29%20%5C%2C%20dz%20%3D%202%5Cpi%20i%20%5Csum_%7B%5Crm%20poles%7D%20%5Cmathrm%7BRes%7D%5C%7Bf%28z%29%5C%7D%20%3D%20%5Cfrac%7B4%5Cpi%7D%7B2%5E%7B7%2F4%7D%7D%20%5Ccos%5Cleft%28%5Cfrac%5Cpi8%5Cright%29)
We also have
![\displaystyle \cos^2\left(\dfrac\pi8\right) = \dfrac{1 + \cos\left(\frac\pi4\right)}2 = \dfrac{2 + \sqrt2}4 \implies \cos\left(\frac\pi8\right) = \dfrac{\sqrt{2+\sqrt2}}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccos%5E2%5Cleft%28%5Cdfrac%5Cpi8%5Cright%29%20%3D%20%5Cdfrac%7B1%20%2B%20%5Ccos%5Cleft%28%5Cfrac%5Cpi4%5Cright%29%7D2%20%3D%20%5Cdfrac%7B2%20%2B%20%5Csqrt2%7D4%20%5Cimplies%20%5Ccos%5Cleft%28%5Cfrac%5Cpi8%5Cright%29%20%3D%20%5Cdfrac%7B%5Csqrt%7B2%2B%5Csqrt2%7D%7D2)
Then the remaining integral is
![\displaystyle \int_0^\infty \frac{t^2+2}{t^4+2t^2+2} \, dt = \frac{4\pi}{2^{7/4}} \cos\left(\frac\pi8\right) = \sqrt{\frac12 + \frac1{\sqrt2}} \, \pi](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%5Cinfty%20%5Cfrac%7Bt%5E2%2B2%7D%7Bt%5E4%2B2t%5E2%2B2%7D%20%5C%2C%20dt%20%3D%20%5Cfrac%7B4%5Cpi%7D%7B2%5E%7B7%2F4%7D%7D%20%5Ccos%5Cleft%28%5Cfrac%5Cpi8%5Cright%29%20%3D%20%5Csqrt%7B%5Cfrac12%20%2B%20%5Cfrac1%7B%5Csqrt2%7D%7D%20%5C%2C%20%5Cpi)
It follows that
![\displaystyle \int_0^{\pi/4} \sqrt{\tan(x) - \tan^2(x)} \, dx = \boxed{\left(\sqrt{\frac12 + \frac1{\sqrt2}} - 1\right) \pi}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cpi%2F4%7D%20%5Csqrt%7B%5Ctan%28x%29%20-%20%5Ctan%5E2%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cboxed%7B%5Cleft%28%5Csqrt%7B%5Cfrac12%20%2B%20%5Cfrac1%7B%5Csqrt2%7D%7D%20-%201%5Cright%29%20%5Cpi%7D)