# Algebra: Solve an equation

**Question**

Solve for x.

(x-2)^2 = (√x) + 2

**Answer**

(x-2)^2 = (√x) + 2

x^2 – 4x +2 – (√x) = 0.

Let (√x) = y. Then,

y^4 – 4y^2 – y +2 = 0

y^2 ( y^2 – 4 ) – ( y – 2 ) = 0

y^2 ( y + 2 ) ( y – 2 ) – ( y – 2 ) = 0

( y – 2 ) ( y^2 [y + 2] – 1 ) = 0.

( y – 2 ) = 0 OR ( y^2 [y + 2] – 1 ) = 0.

So, y = 2. OR ( y^2 [y + 2] – 1 ) = 0. Lets solve for y in the second one…

( y^2 [y + 2] – 1 ) = 0

y^3 + 2 y^2 – 1 = 0

y^3 + y^2 + y^2 – 1 = 0

y^2 [ y + 1 ] + [ y + 1] [ y – 1 ] = 0

[ y + 1 ] [ y^2 + y – 1 ] = 0

[ y + 1 ] = 0 OR [ y^2 + y – 1 ] = 0

So, here y = -1 or [ y^2 + y – 1 ] = 0. As before, lets solve for the ‘second one’.

y^2 + y – 1 = 0 This is a quadratic equation.

Its discriminant = 1^2 – 4 (1) (-1) = 5. So,

y = [-1 +- sqrt(5) ] / 2 which gives two answers namely,

y = – 1.618 OR y = 0.618

Hence we have found the four solutions to our 4th degree equation and these are :

y = ( 2, -1 , – 1.618 , 0.618 )

Since, (√x) = y, thus x = y^2. and thus,

x = ( 2^2, (-1)^2 ,( – 1.618)^2 , ( 0.618)^2 )

Yielding the ‘4’ answers for x as :

x = ( 4 , 1 , 2.6179 , 0.3819 )