1. **Problem statement:** Solve the equation $$x^2 = |2x + 4| + |x - 2|$$ for real values of $x$. 2. **Key idea:** Because of the absolute values, split the problem into intervals based on the points where the expressions inside the absolute values change sign. These points are: - $2x + 4 = 0 \Rightarrow x = -2$ - $x - 2 = 0 \Rightarrow x = 2$ So consider three intervals: $(-\infty, -2)$, $[-2, 2]$, and $(2, \infty)$. 3. **Interval 1: $x < -2$** - $2x + 4 < 0$ so $|2x + 4| = -(2x + 4) = -2x - 4$ - $x - 2 < 0$ so $|x - 2| = -(x - 2) = -x + 2$ Equation becomes: $$x^2 = (-2x - 4) + (-x + 2) = -3x - 2$$ Rewrite: $$x^2 + 3x + 2 = 0$$ Factor: $$(x + 1)(x + 2) = 0$$ Solutions: $$x = -1, -2$$ Check domain $x < -2$: - $x = -2$ is boundary, include it. - $x = -1$ is not in $x < -2$, discard. So valid solution here: $x = -2$. 4. **Interval 2: $-2 \leq x \leq 2$** - $2x + 4 \geq 0$ so $|2x + 4| = 2x + 4$ - $x - 2 \leq 0$ so $|x - 2| = -(x - 2) = -x + 2$ Equation becomes: $$x^2 = (2x + 4) + (-x + 2) = x + 6$$ Rewrite: $$x^2 - x - 6 = 0$$ Factor: $$(x - 3)(x + 2) = 0$$ Solutions: $$x = 3, -2$$ Check domain $-2 \leq x \leq 2$: - $x = -2$ valid - $x = 3$ invalid So valid solution here: $x = -2$ (already found). 5. **Interval 3: $x > 2$** - $2x + 4 > 0$ so $|2x + 4| = 2x + 4$ - $x - 2 > 0$ so $|x - 2| = x - 2$ Equation becomes: $$x^2 = (2x + 4) + (x - 2) = 3x + 2$$ Rewrite: $$x^2 - 3x - 2 = 0$$ Use quadratic formula: $$x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$$ Approximate: - $x_1 = \frac{3 + 4.123}{2} = 3.561$ (valid, since $> 2$) - $x_2 = \frac{3 - 4.123}{2} = -0.561$ (invalid, since $\leq 2$) So valid solution here: $x \approx 3.561$. 6. **Summary of solutions:** - $x = -2$ (from intervals 1 and 2) - $x \approx 3.561$ (from interval 3) 7. **Check solutions in original equation:** - For $x = -2$: $$LHS = (-2)^2 = 4$$ $$RHS = |2(-2) + 4| + |-2 - 2| = |0| + |-4| = 0 + 4 = 4$$ Valid. - For $x \approx 3.561$: $$LHS = (3.561)^2 \approx 12.68$$ $$RHS = |2(3.561) + 4| + |3.561 - 2| = |7.122 + 4| + |1.561| = 11.122 + 1.561 = 12.683$$ Close enough, valid. **Your work had some sign and algebra errors, especially in handling the absolute values and quadratic formula steps.** **Final correct solutions:** $$x = -2, \quad x = \frac{3 + \sqrt{17}}{2}$$