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True or False. The equation below has no solutions. \( \sqrt{x}+\sqrt{x-8}=2 \)

Ask by Reed Fleming. in the United States
Jan 21,2025

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Answer

The equation has no solutions.

Solution

**True.** Let's analyze the equation: \[ \sqrt{x} + \sqrt{x - 8} = 2 \] 1. **Domain Consideration:** - For \(\sqrt{x}\) to be real, \(x \geq 0\). - For \(\sqrt{x - 8}\) to be real, \(x - 8 \geq 0 \Rightarrow x \geq 8\). - Therefore, the domain is \(x \geq 8\). 2. **Evaluating the Equation:** - Let \(y = \sqrt{x}\). Then, \(\sqrt{x - 8} = \sqrt{y^2 - 8}\). - The equation becomes: \[ y + \sqrt{y^2 - 8} = 2 \] - Since \(x \geq 8\), \(y = \sqrt{x} \geq \sqrt{8} \approx 2.828\). - Therefore, \(y \geq 2.828\), which implies: \[ y + \sqrt{y^2 - 8} \geq 2.828 + 0 > 2 \] - This means the left side of the equation is always greater than 2, making the equation impossible to satisfy. Since there is no \(x\) that satisfies the equation within the defined domain, the statement is **True**.

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Extra Insights

Let's unravel this equation! First, we can isolate one of the square roots, let's say \( \sqrt{x} = 2 - \sqrt{x-8} \). Squaring both sides leads us to a more manageable form, but we must keep in mind that squaring can introduce extraneous solutions. Solving that will ultimately reveal whether \( x \) can exist or not. Now, for the values of \( x \), it’s important to check the conditions under which both square roots are defined. Specifically, \( x \) must be greater than or equal to 0, and \( x-8 \) must be greater than or equal to 0. That gives us the condition \( x \geq 8 \) for valid solutions. Fun fact: by substituting values, we find that \( x = 8 \) works, showing that the equation does indeed have a solution—so the answer is False!

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