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Solve for \( x \) \( \begin{array}{l}\sqrt{x+12}=\sqrt{x}+2 \\ x=16 \\ x=2 \\ x=4 \\ \text { no solution }\end{array} \) \( \begin{array}{l}x=2 \\ \text { x }\end{array} \)

Ask by Allan Cole. in the United States
Dec 20,2024

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\( x=4 \)

Solución

Solve the equation \( \sqrt{x+12}=\sqrt{x}+2 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{x+12}=\sqrt{x}+2\) - step1: Find the domain: \(\sqrt{x+12}=\sqrt{x}+2,x\geq 0\) - step2: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{x+12}\right)^{2}=\left(\sqrt{x}+2\right)^{2}\) - step3: Evaluate the power: \(x+12=x+4\sqrt{x}+4\) - step4: Cancel equal terms: \(12=4\sqrt{x}+4\) - step5: Swap the sides: \(4\sqrt{x}+4=12\) - step6: Move the constant to the right side: \(4\sqrt{x}=12-4\) - step7: Calculate: \(4\sqrt{x}=8\) - step8: Divide both sides: \(\sqrt{x}=2\) - step9: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{x}\right)^{2}=2^{2}\) - step10: Evaluate the power: \(x=4\) - step11: Check if the solution is in the defined range: \(x=4,x\geq 0\) - step12: Find the intersection: \(x=4\) - step13: Check the solution: \(x=4\) The solution to the equation \( \sqrt{x+12}=\sqrt{x}+2 \) is \( x=4 \).

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To solve the equation \( \sqrt{x+12}=\sqrt{x}+2 \), start by squaring both sides to eliminate the square roots: \[ x + 12 = (\sqrt{x} + 2)^2 \] Expanding the right side gives: \[ x + 12 = x + 4\sqrt{x} + 4 \] Subtracting \( x \) from both sides simplifies it down to: \[ 12 = 4\sqrt{x} + 4 \] Now, isolate the term with the square root: \[ 8 = 4\sqrt{x} \] Dividing both sides by 4: \[ 2 = \sqrt{x} \] Finally, squaring both sides yields: \[ x = 4 \] Now check the solution by substituting \( x = 4 \) back into the original equation: \[ \sqrt{4 + 12} = \sqrt{4} + 2 \implies \sqrt{16} = 2 + 2 \implies 4 = 4 \] That checks out, so the solution is indeed \( x = 4 \). --- Did you know that the method of squaring both sides of an equation can lead to "extraneous" solutions? Always remember to check your solutions by substituting them back into the original equation, just like we did here! In real-world applications, recognizing and solving square root equations is crucial, especially in fields like engineering and physics where variables may represent physical quantities. For example, this technique can help in calculating distances or time taken in motion problems!

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