How do you solve #(1/x)+1/(x+3)=1/4# using the quadratic formula?

1 Answer
May 3, 2016

Answer:

#color(blue)(x=6.772" and" -1.772)#

Explanation:

Consider the left side

Common denominator is #x(x+3)#

So we have:

#((x+3)+x)/(x(x+3))=1/4#

#(2x+3)/(x^2+3x)=1/4#

Multiply both sides by #(x^2+3x)#

#(2x+3)xx(x^2+3x)/(x^2+3x)=(x^2+3x)/4#

but #(x^2+3x)/(x^2+3x)=1#

#(2x+3)=(x^2+3x)/4#

Multiply both sides by 4

#4(2x+3)=x^2+3x#

#8x+12=x^2+3x#

Subtract #8x# and 12 from both sides

#x^2+3x-8x-12=0#

#color(brown)(x^2-5x-12=0)#
#color(brown)("Now we can use the formula")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Standard form #y=ax^2+bx+c#
where #a=1" ; "b=-5" ; "c=-12#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(5+-sqrt((-5)^2-4(1)(-12)))/(2(1)) #

#x=(5+-sqrt(25+48))/2 #

#x=(5+-sqrt(73))/2" " # Note that 73 is a prime number

#color(blue)(x=6.772" and" -1.772)#

Tony B