Answer:
2
Step-by-step explanation:
So I'm going to use vieta's formula.
Let u and v the zeros of the given quadratic in ax^2+bx+c form.
By vieta's formula:
1) u+v=-b/a
2) uv=c/a
We are also given not by the formula but by this problem:
3) u+v=uv
If we plug 1) and 2) into 3) we get:
-b/a=c/a
Multiply both sides by a:
-b=c
Here we have:
a=3
b=-(3k-2)
c=-(k-6)
So we are solving
-b=c for k:
3k-2=-(k-6)
Distribute:
3k-2=-k+6
Add k on both sides:
4k-2=6
Add 2 on both side:
4k=8
Divide both sides by 4:
k=2
Let's check:
:
![3x^2-(3\cdot 2-2)x-(2-6)](https://tex.z-dn.net/?f=3x%5E2-%283%5Ccdot%202-2%29x-%282-6%29)
![3x^2-4x+4](https://tex.z-dn.net/?f=3x%5E2-4x%2B4)
I'm going to solve
for x using the quadratic formula:
![\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}](https://tex.z-dn.net/?f=%5Cfrac%7B4%5Cpm%20%5Csqrt%7B%28-4%29%5E2-4%283%29%284%29%7D%7D%7B2%283%29%7D)
![\frac{4\pm \sqrt{16-16(3)}}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B4%5Cpm%20%5Csqrt%7B16-16%283%29%7D%7D%7B6%7D)
![\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B4%5Cpm%20%5Csqrt%7B16%7D%5Csqrt%7B1-%283%29%7D%7D%7B6%7D)
![\frac{4\pm 4\sqrt{-2}}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B4%5Cpm%204%5Csqrt%7B-2%7D%7D%7B6%7D)
![\frac{2\pm 2\sqrt{-2}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Cpm%202%5Csqrt%7B-2%7D%7D%7B3%7D)
![\frac{2\pm 2i\sqrt{2}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Cpm%202i%5Csqrt%7B2%7D%7D%7B3%7D)
Let's see if uv=u+v holds.
![uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}](https://tex.z-dn.net/?f=uv%3D%5Cfrac%7B2%2B2i%5Csqrt%7B2%7D%7D%7B3%7D%20%5Ccdot%20%5Cfrac%7B2-2i%5Csqrt%7B2%7D%7D%7B3%7D)
Keep in mind you are multiplying conjugates:
![uv=\frac{1}{9}(4-4i^2(2))](https://tex.z-dn.net/?f=uv%3D%5Cfrac%7B1%7D%7B9%7D%284-4i%5E2%282%29%29)
![uv=\frac{1}{9}(4+4(2))](https://tex.z-dn.net/?f=uv%3D%5Cfrac%7B1%7D%7B9%7D%284%2B4%282%29%29)
![uv=\frac{12}{9}=\frac{4}{3}](https://tex.z-dn.net/?f=uv%3D%5Cfrac%7B12%7D%7B9%7D%3D%5Cfrac%7B4%7D%7B3%7D)
Let's see what u+v is now:
![u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}](https://tex.z-dn.net/?f=u%2Bv%3D%5Cfrac%7B2%2B2i%5Csqrt%7B2%7D%7D%7B3%7D%2B%5Cfrac%7B2-2i%5Csqrt%7B2%7D%7D%7B3%7D)
![u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}](https://tex.z-dn.net/?f=u%2Bv%3D%5Cfrac%7B2%7D%7B3%7D%2B%5Cfrac%7B2%7D%7B3%7D%3D%5Cfrac%7B4%7D%7B3%7D)
We have confirmed uv=u+v for k=2.
100
75 times 1.333333333333333 equals 100
Make a change of coordinates:
![u(x,y)=xy](https://tex.z-dn.net/?f=u%28x%2Cy%29%3Dxy)
![v(x,y)=\dfrac xy](https://tex.z-dn.net/?f=v%28x%2Cy%29%3D%5Cdfrac%20xy)
The Jacobian for this transformation is
![\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20J%3D%5Cbegin%7Bbmatrix%7D%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%26%5Cdfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D%5C%5C%5C%5C%5Cdfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D%26%5Cdfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7Dy%26x%5C%5C%5C%5C%5Cdfrac1y%26-%5Cdfrac%20x%7By%5E2%7D%5Cend%7Bbmatrix%7D)
and has a determinant of
![\det\mathbf J=-\dfrac{2x}y](https://tex.z-dn.net/?f=%5Cdet%5Cmathbf%20J%3D-%5Cdfrac%7B2x%7Dy)
Note that we need to use the Jacobian in the other direction; that is, we've computed
![\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}](https://tex.z-dn.net/?f=%5Cmathbf%20J%3D%5Cdfrac%7B%5Cpartial%28u%2Cv%29%7D%7B%5Cpartial%28x%2Cy%29%7D)
but we need the Jacobian determinant for the reverse transformation (from
![(x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29)
to
![(u,v)](https://tex.z-dn.net/?f=%28u%2Cv%29)
. To do this, notice that
![\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28x%2Cy%29%7D%7B%5Cpartial%28u%2Cv%29%7D%3D%5Cdfrac1%7B%5Cdfrac%7B%5Cpartial%28u%2Cv%29%7D%7B%5Cpartial%28x%2Cy%29%7D%7D%3D%5Cdfrac1%7B%5Cmathbf%20J%7D)
we need to take the reciprocal of the Jacobian above.
The integral then changes to
![\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7B%5Cmathcal%20W_%7B%28x%2Cy%29%7D%7De%5E%7Bxy%7D%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D%5Ciint_%7B%5Cmathcal%20W_%7B%28u%2Cv%29%7D%7D%5Cdfrac%7Be%5Eu%7D%7B%7C%5Cdet%5Cmathbf%20J%7C%7D%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)