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

Hint: there is a tricky way to solve it !

tricky way

The two solutions for x :
(-1 - sqrt(13))/2 ,   (1 + sqrt(17))/2

Notice that the LHS and RHS are each a parabola, though the RHS is rotated 90 degrees.  These 2 parabolas will have 4 intersection points, but due to the square root sign, y >= 0, so 2 of the 4 solutions will be rejected.

Let y = RHS and y = LHS
y = x^2 - 4
y^2 = x + 4
so there is a symmetric system:
y = x^2 - 4  eqn 1
x = y^2 - 4  eqn 2

Subtract:
(y-x) = x^2 - y^2 = (x-y)(x+y)
(y-x) - (x-y)(x+y) = 0
(y-x) + (y-x)(x+y) = 0
(y-x)(x+y+1) = 0
case 1:  y = -(x+1)   or   
case 2:  y = x
plug into eqn 2

case 1: y= -(x+1)
x = x^2 + 2x + 1 - 4
x^2 + x - 3 = 0
(x,y) = ( (-1-sqrt(13))/2, -(1-sqrt(13))/2 )
(x,y) = ( (-1+sqrt(13))/2, -(1+sqrt(13))/2 )  rejected

case 2:  y=x
x = x^2 - 4
x^2 - x - 4 = 0
(x,y) = ( (1+sqrt(17))/2, (1+sqrt(17))/2 )
(x,y) = ( (1-sqrt(17))/2, (1-sqrt(17))/2 )  rejected

---
The non-tricky way is to square both sides, getting a quartic equation where the x^4 coefficient is 1 and the x^3 coefficient is 0:
x^4 - 8x^2 - x + 12 = 0
Since there is no x^3 term, the quartic can be factored as:
(x^2 + ax + b)(x^2 - ax + c)  and we equate like terms to get:
-8 = b+c-a^2
-1 = ac-ab
12 = bc
This is solved with (a,b,c) = (1,-3,-4) giving:
(x^2 + x - 3)(x^2 - x - 4), the same two quadratics as above.

Posted by Larry on 2023-03-20 11:32:45