# `4x+4root(x))(x-2)=-(2-X)^2` Solve

pramodpandey | Student

`4x+4sqrt(x)(x-2)=-(2-x)^2`

`4x+4sqrt(x)(x-2)+(2-x)^2=0`

`2^2(sqrt(x))^2-2xx2sqrt(x)(2-x)+(2-x)^2=0`

we know `(a-b)^2=a^2-2ab+b^2=(a-b)^2`

here      `a=2sqrt(x)`

`b=(2-x)`

`2^2(sqrt(x))^2-2xx2sqrt(x)(2-x)+(2-x)^2=(2sqrt(x)-(2-x))^2`

`(2sqrt(x)-(2-x))^2=0`

`2sqrt(x)-(2-x)=0`

`2sqrt(x)=2-x`

Squaring both side

`(2sqrt(x))^2=(2-x)^2`

`4x=4-4x+x^2`

`x^2-8x-4=0`

`x=(8+-sqrt(8^2+4xx(-4)))/2`

`x=(8+-sqrt(80))/2`

`x=4+-sqrt(20)`

`x=4-sqrt(20)<0`  is not possible because of the term `sqrt(x)`  in original equation. `sqrt(4-sqrt(20))`  is complex no.

So  x=`4+sqrt(20)`

oldnick | Student

`4x +(x -2)4sqrt(x)= -(2-x)^2`

adding both sides by  `(2 - x)^2`  we get:

`4x + 4(x-2)sqrt(x)+(2-x)^2= -(2-x)^2 +(2-x)^2`

So erasing terms do like:

`4x + 4(x -2)sqrt(x) +(2-x)^2= 0`

Now we now that: `(2-x)^2=(x-2)^2`

`4x +4(x-2)sqrt(x) +(x -2)^2= 0`

rewriting in this terms:

`(2sqrt(x))^2 + 2 (x-2) 2sqrt(x) +(x-2)^2 =0`

it shows at once be  the square of `2sqrt(x) + x - 2`

so that:

`(2sqrt(x)+x -2 )^2= 0`

`2sqrt(x) + x -2 =0`

subract from both sides  `x - 2`

`2sqrt(x) + (x - 2) - (x - 2)= -(x - 2)`

semplifyng:

`2sqrt(x) = -(x -2)`

squariing:

`4x=x^2-4x + 4`

subract now both sides `4x`

`4x -4x=x^2 -4x -4x +4`

so:

`x^2 - 8x +4 = 0`

adding both sides 12 we get:

`x^2-8x + 4 + 12= 12`

and:

`x^2-8x+16=12`

left side sound as  the x - 4 square:

`(x-4)^2=12`

extratctin square root:

`x-4=` `+-` `2sqrt(3)`

so that we have got two solution:

`x= 4 +2sqrt(3)`  and  `x=4 - 2sqrt(3)`