Solve Equation

[Guest]

(x - 3b) / c = 9(c - b) / x

Solve.


x^2 - 3bx = 9c^2 - bc Cross multiplied.

Not sure how to proceed.

[Baltuilhe]

Good afternoon!

[tex]\frac{x-3b}{c}=\frac{9(c-b)}{x}\\
x(x-3b)=9c(c-b)\\
x^2-3bx=9c^2-9bc\\
x^2-3bx-9c^2+9bc=0\\
\Delta=(-3b)^2-4(1)(-9c^2+9bc)\\
\Delta=9b^2+36c^2-36bc=9\left(b^2-4bc+4c^2\right)=9\left(b-2c\right)^2\\
x=\frac{-(-3b)\pm\sqrt{\Delta}}{2(1)}\\
x=\frac{3b\pm 3(b-2c)}{2}\\
x'=\frac{3b+3b-6c}{2}=3(b-c)\\
x''=\frac{3b-3b+6c}{2}=3c[/tex]

Hope to Have Helped (3H)

[Guest]

I don't understand your two steps that begin with the triangles.

[Baltuilhe]

Good afternoon!

It's the solution to 2nd degree equation:
[tex]ax^2+bx+c=0\Rightarrow
\begin{cases}\Delta=b^2-4ac\\
x=\frac{-b\pm\sqrt{\Delta}}{2a}\Rightarrow\begin{cases}x'=\frac{-b+\sqrt{\Delta}}{2a}\\x''=\frac{-b-\sqrt{\Delta}}{2a}\end{cases}
\end{cases}[/tex]

See?

[Guest]

Good Afternoon!

Ok, Thanks again.