How do you solve #10/(x^2-25) - 1/10 = 1/(x - 5)#?

1 Answer
Apr 15, 2018

Answer:

#x_1=5#
#x_2=-15#

Explanation:

#x^2-25=(x-5)*(x+5) # so we have

#10/((x+5)*(x-5))-1/10=1/(x-5)#

multiply both sides by #10(x+5)(x-5)#

#10(x+5)(x-5)10/[(x+5)(x-5)]-10(x+5)(x-5)1/10=10(x+5)(x-5)1/(x-5)#

#10color(red)cancel(x+5)color(red)cancel(x-5)10/[color(red)cancel(x+5)color(red)cancel(x-5)]-color(blue)cancel10(x+5)(x-5)1/color(blue)cancel10=10(x+5)color(green)cancel(x-5)1/color(green)cancel(x-5)#

#100-(x^2-25)=10x+50#

#100-x^2+25-10x-50=0#

#-x^2-10x+75=0#

#x^2+10x-75=0#

now use the formula #x_(1//2)=(-b+-sqrt(b^2-4ac))/(2a)#
#a=1#
#b=10#
#c=-75#

#x_(1//2)=(-10+-sqrt(100-4(-75)))/(2)#
#x_(1//2)=(-10+-sqrt(400))/(2)#
#x_(1//2)=(-10+-20)/2#
#x_1=(-10+20)/2=10/2=5#
#x_2=(-10-20)/2=-30/2=-15#