Question

\ sqrt {x ^ 2 + 4x + 7} = \ sqrt {x ^ 2-2x + 4

Answer 1

First note the domain of the expressions on either side of the equation,

$\sqrt{x^2 + 4x + 7} = \sqrt{x^2 - 2x + 4}$

√x is defined only for x ≥ 0, so we need to have

$x^2 + 4x + 7 \ge 0 \\\\ (x^2+4x+4) + 3 \ge 0 \\\\ (x+2)^2 \ge -3$

and

$x^2 - 2x + 4 \ge 0 \\\\ (x^2 - 2x + 1) + 3 \ge 0 \\\\ (x-1)^2 \ge -3$

but x ² ≥ 0 for all real x, so both conditions will always be satisfied.

Back to the equation - take the square of both sides and solve for x :

$x^2 + 4x + 7 = x^2 - 2x + 4 \\\\ 4x + 7 = -2x + 4 \\\\ 6x = -3 \\\\ \boxed{x=-\dfrac12}$