How do you solve the rational equation #(x^2-7)/(x-4)=0#?

2 Answers
Jan 11, 2016

Answer:

# x = ±sqrt7 #

Explanation:

for this rational function to equate to zero means that the numerator must be zero as the denominator ≠ 0

# rArr x^2 - 7 = 0 rArr x^2 = 7 rArr x = ±sqrt7 #

Jan 11, 2016

Answer:

#x=+-sqrt(7)#

Strong guidance given on manipulation!

Explanation:

Given: #color(brown)(color(white)(....) (x^2-7)/(x-4)=0)#

Multiply both sides by #color(blue)((x-4))#

#color(brown)((x^2-7)/(x-4)color(blue)(xx(x-4))=0xxcolor(blue)((x-4))#

#(x^2-7) xx(x-4)/(x-4)=0#

But #(x-4)/(x-4)" is another way of writing " 1# giving:

#(x^2-7) xx1=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Add #color(blue)(7)# to both sides

#color(brown)( x^2-7 color(blue)(+7) =0color(blue)(+7)#

#x^2=7 #
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

so #color(white)(...) x= sqrt(x^2) = +-sqrt(7) #

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The square root has to be #+-# because:

#x^2=( + sqrt(7))xx (+sqrt(7) )-> "positive "x^2#

And

#x^2= (-sqrt(7))xx(-sqrt(7))-> "positive "x^2#