How do you solve?

#root(4)(194-x)+root(4)(x)=4# ?

1 Answer
Apr 6, 2018

#color(red)(x=97+-56sqrt3#
#color(red)(a^4+b^4=(a^2+b^2)^2-2a^2b^2#

Explanation:

Here,

#root(4)(194-x)+root(4)x=4#

Let, #color(red)a=root(4)(194-x)# #andcolor(red) b=root(4)x=>color(red)(a^4)=194-x andcolor(red)( b^4)=x#

#:.color(blue)(a+b=4...to(I) and a^4+b^4=194...to(II)#

Squaring #(I)=>(a+b)^2=4^2=>a^2+2ab+b^2=16#

i.e. #color(blue)(a^2+b^2=16-2ab...to(III)#

From #(II)toa^4+b^4=194#,

#=>color(brown)((a^2+b^2)^2-2a^2b^2=194#

#=>(16-2ab)^2-2a^2b^2=194...to#using #(III)#

#=>256-64ab+4a^2b^2-2a^2b^2=194#

#=>2a^2b^2-64ab+256=194#

#=>a^2b^2-32ab+128=97#

#=>a^2b^2-32ab+256=97+128=225#

#=>(ab-16)^2=15^2#

#=>ab-16=+-15#

#=>ab=16+-15=>color(blue)(ab=1 or ab=31#

If #ab=1,then ,a=1/b to# #""color(red)"Please see the comment below"#

From #(I)to1/b+b=4=>1+b^2=4b#

#=>b^2-4b+1=0=>b^2-4b+4=3=>(b-2)^2=(sqrt3)^2#

#=>b-2=+-sqrt3=>b=2+-sqrt3#

#b^2=(2+-sqrt3)^2=4+-4sqrt3+3=7+-4sqrt3#

#b^4=(7+-4sqrt3)^2=49+-56sqrt3+48#

#color(red)(x=97+-56sqrt3#

Note:If #ab=31=>a=31/b#

From #(I)to31/b+b=4=>b^2-4b+31=0#

#triangle=16-124<0=>color(blue)(ab!=31#