How do you solve #x^{1/ 2} - 4x^{1/ 4} = - 3#?

1 Answer
Sep 23, 2016

#x=81#
Or
#x=1#

Explanation:

Take #color(blue)(u=x^(1/4))#

#then#
#u^2=x^((1/4)^2)#
#u^2=x^(2/4)#
#color(red)(u^2=x^(1/2))#

Let us substitute the calculated expressions above in the given expression we have:

#color(red)(x^(1/2))-4color(blue)(x^(1/4))=-3#
#color(red)(u^2)-4color(blue)(u)=-3#
#color(red)(u^2)-4color(blue)(u)+3=0#

Here we can check if we can find two real numbers #a# and #b# such that #a+b=-4# and #a*b=3#
Yes, we can here in the above expression #a=-3# and #b=-1#

#color(red)(u^2)-4color(blue)(u)+3=0#
#color(green)((u-3))color(purple)((u-1))=0#
So,

#color(green)(u-3)=0 #
then
#color(blue)u=3#
#color(blue)(x^(1/4))=3#
#color(blue)((x^(1/4)))^4=3^4#
#x=81#

Or
#(u-1)=0 #
then
#color(blue) u=1#
#color(blue)(x^(1/4)=1#
#color(blue)((x^(1/4)))^4=1^4#
#x=1#