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