Fractional Equations #2

Algebra Level 4

Find the sum of all possible real values of $x$ that satisfy the equation:

$\frac{x^2+4x}{x-1}+\frac{72x-72}{x^2+4x}-18=0.$

2 solutions

U Z
Oct 26, 2014

$y =\frac{ x^2 + 4x}{ x -1}$

$y + \frac{72}{y} - 18 = 0$

$y^{2} - 18y + 72 =0$

$y =6 , 12$

$\frac{x^{2} + 4x}{ x -1} = 6 , \frac{x^{2} + 4x}{ x -1} = 12$

for y=6 the perfect square comes negative therefore y= 6 is not possible

when y =12 we get $x = 6 , 2$

Awesome explanation!

William Isoroku - 6 years, 5 months ago

William Isoroku
Dec 20, 2014

You can also solve this by graphing and find the x-intercepts.