Question - Solving Equations with Absolute Values and Radicals
Solution:
The image shows two equations that we are asked to solve over the real number set $$\mathbb{R}$$:1) $$x^2 - 2x - 2|x - 1| - 2 = 0$$2) $$(x + 2022)(\sqrt{x^2 + 2021} - 1) = 2021$$Let's solve them one by one.For the first equation:To solve this equation, we must consider the absolute value function which splits the domain into two parts, depending on the sign of the expression inside the absolute value function.Absolute value function rules:- If $$x - 1 \geq 0$$ (which means if $$x \geq 1$$), then $$|x - 1| = x - 1$$.- If $$x - 1 < 0$$ (which means if $$x < 1$$), then $$|x - 1| = -(x - 1) = -x + 1$$.So, we need to consider two cases:Case 1: $$x \geq 1$$If $$x \geq 1$$, then $$|x - 1| = x - 1$$, so our equation becomes $$x^2 - 2x - 2(x - 1) - 2 = 0$$, which simplifies to:$$x^2 - 2x - 2x + 2 - 2 = 0$$$$x^2 - 4x = 0$$$$x(x - 4) = 0$$This gives us two solutions: $$x = 0$$ and $$x = 4$$. However, since $$x = 0$$ is not in the domain where $$x \geq 1$$, we discard it. So the valid solution in this case is $$x = 4$$.Case 2: $$x < 1$$If $$x < 1$$, then $$|x - 1| = -x + 1$$, so our equation becomes $$x^2 - 2x - 2(-x + 1) - 2 = 0$$, which simplifies to:$$x^2 - 2x + 2x - 2 - 2 = 0$$$$x^2 - 4 = 0$$$$x^2 = 4$$Taking the square root of both sides, we get $$x = \pm 2$$. Since $$x$$ must be less than $$1$$, we discard $$x = 2$$ and only $$x = -2$$ is within the valid domain.So, the solution to equation 1) based on the split due to the absolute value is $$x = 4$$ and $$x = -2$$.Now for the second equation:The second equation is $$(x + 2022)(\sqrt{x^2 + 2021} - 1) = 2021$$. To solve this, we can set each factor equal to zero.If $$x + 2022 = 0$$, then $$x = -2022$$.For the other factor $$\sqrt{x^2 + 2021} - 1 = 0$$, we add $$1$$ to both sides and then square both sides:$$\sqrt{x^2 + 2021} = 1$$$$x^2 + 2021 = 1$$$$x^2 = 1 - 2021$$$$x^2 = -2020$$Since $$x^2$$ cannot be negative in the set of real numbers, there are no real solutions from the second factor. Therefore, the only solution from the second equation is $$x = -2022$$.To summarize:1) $$x = 4$$ and $$x = -2$$2) $$x = -2022$$
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