Know more about quadratic equations http://en.wikipedia.org/wiki/Quadratic_equation
Question:
(x + 2) / (x - 3) = 42 (x + 7)
Solution:
(x +2 ) (x + 7) = 42 (x - 3)
x² + 7x + 2x + 14 = 42x - 126
x² + 9x + 14 = 42x - 126
x² + 9x + 14 - 42x + 126 = 0
x² - 33x + 140 = 0
x² - 28x - 5x + 140 = 0
Taking out common factors
x(x - 28) -5(x - 28) = 0
(x - 5) (x - 28) = 0
x -5 = 0
OR
x - 28 = 0
Either
x = 5
OR
x = 28
For more help on solving quadratic equations.
![f(x,y) = xe^{x^2y}\\\frac{\partial f}{\partial x}=? And \frac{\partial f}{\partial y}=?\\\frac{\partial f}{\partial x}_{(1,2)}=? And \frac{\partial f}{\partial y}_{(1,ln 2)}=?\\ f(x,y) = xe^{x^2y}<br />\\Consider \frac{\partial e^{x^2y}}{\partial x}\\\frac{\partial e^u}{\partial x} and u=x^2y\\\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\ as u=x^2y\\\frac{\partial e^u}{\partial u}=e^u=e^{x^2y}----------(a)\\\frac{\partial u}{\partial x}=2xy-------(b)\\ By Chain Rule\\\frac{\partial e^{x^2y}}{\partial x}=\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\=e^{x^2y} . 2xy where a = e^{x^2y} and b=2xy\\\frac{\partial f}{\partial x}=x \frac{\partial e^{x^2y}}{\partial x}+e^{x^2y}\frac{\partial x}{\partial x}\\=xe^{x^2y}(2xy)+e^{x^2y}(1)\\=e^{x^2y}[2x^2y+1]\\\frac{\partial f}{\partial x}=xe^{x^2y}(x^2)\\=x^3e^{x^2y}\\Therefore\frac{\partial f}{\partial x}_{(1,ln2)}=e^{1ln2}(i^2.ln2)=e^{ln2}=2ln2\\\frac{\partial f}{\partial y}_{(1,ln2)}=1^3 e^{1^2ln(2)}=e^{ln(2)}=2<br />](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDA9G1EQoYK18sd1GibXDdHa8cpFAP50o9rwzNzbmdv5Xa2-Uz0wNPFBXtUc9AUkcbzkYKYksefcitwhs8g46nGNw6BAgNsijf7mJuiJwxi4syAbe7q8miLe71y4VTOX0_OET_ktTX4cM/s1600-h/finaldydx.JPG)
