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Ludmilka [50]
3 years ago
7

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

Mathematics
1 answer:
slavikrds [6]3 years ago
6 0

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

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

√<em>x</em> is defined only for <em>x</em> ≥ 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 <em>x</em> ² ≥ 0 for all real <em>x</em>, so both conditions will always be satisfied.

Back to the equation - take the square of both sides and solve for <em>x</em> :

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

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