504,726 views
7 votes
7 votes
\ sqrt {x ^ 2 + 4x + 7} = \ sqrt {x ^ 2-2x + 4

\ sqrt {x ^ 2 + 4x + 7} = \ sqrt {x ^ 2-2x + 4-example-1
User Dwj
by
2.7k points

1 Answer

10 votes
10 votes

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


√(x^2 + 4x + 7) = √(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=-\frac12}

User Mnist
by
2.7k points