Answer to Question #164304 in Math for M

Question

Solve the following equation:

x Ix-5I = 20x+3

List the four possible roots in increasing order and indicate if is a EXTRANEOUS root or just a ROOT.

Answer 1

$x|x-5|=20x+3$

$x\geq5, |x-5|=x-5$

$x(x-5)=20x+3$

$x^2-5x-20x-3=0$

$x^2-25x-3=0$

$D=b^2-4ac=(-25)^2-4(1)(-3)=637$

$x_1=\dfrac{25-\sqrt{637}}{2(1)}=\dfrac{25-7\sqrt{13}}{2}$

$x_2=\dfrac{25+\sqrt{637}}{2(1)}=\dfrac{25+7\sqrt{13}}{2}$

$x<5, |x-5|=-(x-5)$

$-x(x-5)=20x+3$

$x^2-5x+20x+3=0$

$x^2+15x+3=0$

$D=b^2-4ac=(15)^2-4(1)(3)=213$

$x_1=\dfrac{-15-\sqrt{213}}{2(1)}=\dfrac{-15-\sqrt{213}}{2}$

$x_2=\dfrac{-15+\sqrt{213}}{2(1)}=\dfrac{-15+\sqrt{213}}{2}$

List the four possible roots in increasing order 

$\dfrac{-15-\sqrt{213}}{2}, \dfrac{-15+\sqrt{213}}{2},$

$\dfrac{25-7\sqrt{13}}{2}, \dfrac{25+7\sqrt{13}}{2}$

$x=\dfrac{-15-\sqrt{213}}{2}$ just root

$x=\dfrac{-15+\sqrt{213}}{2}$ just root

$x=\dfrac{25-7\sqrt{13}}{2}$ extraneous root

$x=\dfrac{25+7\sqrt{13}}{2}$ just root