how to solve equation with square root

[Albert Chan]

(01-03-2020 07:08 AM)teerasak Wrote:  When using solve(Sqrt(x^2+x-2) + Sqrt(x^2+x-30) = Abs(Sqrt(2*x^2+2*x-32)),x)

I expect it should fiter answer with intermediate complex number result, but it does not.

Think of it if you solve this manually ...

LHS and RHS both non-negative. Square both side:

\((x^2 + x - 2) + (x^2 + x - 30) + 2 \sqrt{(x^2 + x - 2) (x^2 + x - 30)} = 2x^2 + 2x -32\)

\(2 \sqrt{(x^2 + x - 2) (x^2 + x - 30)} = 0\)

Again, both side is non-negative, square again:

\(4(x^2 + x - 2) (x^2 + x - 30) = 4(x+2)(x-1)(x+6)(x-5) = 0 \)

Note that above does not involve complex numbers.

To remove "complex intermediates", we need RHS = \(2x^2 + 2x -32 ≥ 0\)
With this extra constraint, we are left with 2 roots, x = -6, +5