How do you solve?

root(4)(194-x)+root(4)(x)=44194x+4x=4 ?

1 Answer
Apr 6, 2018

color(red)(x=97+-56sqrt3x=97±563
color(red)(a^4+b^4=(a^2+b^2)^2-2a^2b^2a4+b4=(a2+b2)22a2b2

Explanation:

Here,

root(4)(194-x)+root(4)x=44194x+4x=4

Let, color(red)a=root(4)(194-x)a=4194x andcolor(red) b=root(4)x=>color(red)(a^4)=194-x andcolor(red)( b^4)=xandb=4xa4=194xandb4=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...tousing (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