Consider the equation
\[
\sqrt{x^{2 } - 2x + 1} = x + 2
\]
and the equation which arises from this equation by squaring both sides of the
equation, i.e. the equation
\[
\left (\sqrt{x^{2 } - 2x + 1}\right )^{2} = (x + 2)^{2}.
\]
Identify a true statement.
Both equations are equivalent only if
\(x\geq - 2\).
Both equations are equivalent.
Both equations are equivalent only if
\(x\leq - 2\).
Removing radical in an equation by squaring both sides may enrich the set of
solutions of this equation and checking the solutions of the new equation in the
original equation may be necessary. Identify a correct conclusion in the particular
case of the following equation.
\[
-\sqrt{x^{2 } - 2x + 1} = x
\]
If we look for the solution in the set
\(\mathbb{R}^{-}\), then
squaring both sides of the equation gives an equivalent equation. The checking of the
solution is not necessary.
If we look for the solution in the set
\(\mathbb{R}^{+}\), then
squaring both sides of the equation gives an equivalent equation. The checking of the
solution is not necessary.
If we look for the solution in the set
\(\mathbb{R}\), then
squaring both sides of the equation gives an equivalent equation. The checking of the
solution is not necessary.