Question

solve |x² - 4| = 2 - x algebraically.

Explanation:

Step 1: Analyze the domain from the right-hand side

The right-hand side is \(2 - x\). For the equation \(|x^{2}-4| = 2 - x\) to be valid, the right-hand side must be non - negative (since the absolute value \(|A|\geq0\) for any real number \(A\)). So we have the inequality \(2 - x\geq0\), which implies \(x\leq2\).

Step 2: Case 1: \(x^{2}-4\geq0\)

If \(x^{2}-4\geq0\), then \(|x^{2}-4|=x^{2}-4\). The inequality \(x^{2}-4\geq0\) can be factored as \((x - 2)(x + 2)\geq0\). The solution to this inequality is \(x\leq - 2\) or \(x\geq2\). But from the domain of the right - hand side, we know \(x\leq2\). So the intersection of \(x\leq - 2\) or \(x\geq2\) and \(x\leq2\) is \(x\leq - 2\) or \(x = 2\).

The equation becomes \(x^{2}-4=2 - x\). Rearranging this equation to standard quadratic form: \(x^{2}+x-6 = 0\). Factoring the quadratic: \((x + 3)(x - 2)=0\). The solutions are \(x=-3\) or \(x = 2\). Now we check these solutions against the condition \(x\leq - 2\) or \(x = 2\). For \(x=-3\), \(-3\leq - 2\) (valid). For \(x = 2\), \(x = 2\) (valid).

Step 3: Case 2: \(x^{2}-4\lt0\)

If \(x^{2}-4\lt0\), then \(|x^{2}-4|=-(x^{2}-4)=-x^{2}+4\). The inequality \(x^{2}-4\lt0\) can be factored as \((x - 2)(x + 2)\lt0\), and its solution is \(-2\lt x\lt2\).

The equation becomes \(-x^{2}+4=2 - x\). Rearranging to standard quadratic form: \(x^{2}-x - 2=0\). Factoring the quadratic: \((x - 2)(x+1)=0\). The solutions are \(x = 2\) or \(x=-1\). Now we check these solutions against the condition \(-2\lt x\lt2\). For \(x = 2\), \(2\) is not in the interval \(-2\lt x\lt2\) (so we discard \(x = 2\)). For \(x=-1\), \(-1\) is in the interval \(-2\lt x\lt2\) (valid).

Step 4: Combine the solutions

From Case 1, we have \(x=-3\) and \(x = 2\). From Case 2, we have \(x=-1\). Now we need to verify all solutions in the original equation:

• For \(x=-3\): Left - hand side \(|(-3)^{2}-4|=|9 - 4| = 5\). Right - hand side \(2-(-3)=5\). So \(x=-3\) is a valid solution.
• For \(x = 2\): Left - hand side \(|2^{2}-4|=|4 - 4| = 0\). Right - hand side \(2 - 2=0\). So \(x = 2\) is a valid solution.
• For \(x=-1\): Left - hand side \(|(-1)^{2}-4|=|1 - 4| = 3\). Right - hand side \(2-(-1)=3\). So \(x=-1\) is a valid solution.

Answer:

The solutions of the equation \(|x^{2}-4|=2 - x\) are \(x=-3\), \(x=-1\), and \(x = 2\).