Question

what is the solution to the equation \\(\sqrt{2n + 28} - 4\sqrt{n} = 0\\)?
\\(\bigcirc\\ n = 2\\)
\\(\bigcirc\\ n = 4\\)
\\(\bigcirc\\ n = 7\\)
\\(\bigcirc\\ n = 14\\)

Explanation:

Step1: Isolate the square root term

We start with the equation \(\sqrt{2n + 28}-4\sqrt{n}=0\). Add \(4\sqrt{n}\) to both sides to get \(\sqrt{2n + 28}=4\sqrt{n}\).

Step2: Square both sides

Squaring both sides of the equation \(\sqrt{2n + 28}=4\sqrt{n}\), we use the property \((\sqrt{a})^2=a\) (for \(a\geq0\)). So \((\sqrt{2n + 28})^2=(4\sqrt{n})^2\), which simplifies to \(2n + 28 = 16n\).

Step3: Solve for \(n\)

Subtract \(2n\) from both sides of \(2n+28 = 16n\): \(28=16n - 2n\), so \(14n=28\). Then divide both sides by \(14\): \(n=\frac{28}{14} = 2\). But we need to check if this solution is valid by plugging it back into the original equation.

Step4: Check the solution

For \(n = 2\), the left - hand side of the original equation is \(\sqrt{2\times2 + 28}-4\sqrt{2}=\sqrt{4 + 28}-4\sqrt{2}=\sqrt{32}-4\sqrt{2}=4\sqrt{2}-4\sqrt{2}=0\), which matches the right - hand side. Wait, but let's check the other options just in case we made a mistake.
Wait, if we made a mistake in step 2, let's re - do step 2. Wait, \((4\sqrt{n})^2=16n\), and \((\sqrt{2n + 28})^2 = 2n+28\). So \(2n + 28=16n\), \(14n = 28\), \(n = 2\). But let's check \(n = 2\) in the original equation: \(\sqrt{2\times2+28}-4\sqrt{2}=\sqrt{32}-4\sqrt{2}=4\sqrt{2}-4\sqrt{2}=0\). Now check \(n = 4\): \(\sqrt{2\times4 + 28}-4\sqrt{4}=\sqrt{8 + 28}-8=\sqrt{36}-8=6 - 8=- 2
eq0\). For \(n = 7\): \(\sqrt{2\times7+28}-4\sqrt{7}=\sqrt{14 + 28}-4\sqrt{7}=\sqrt{42}-4\sqrt{7}\approx6.48-10.58=-4.1
eq0\). For \(n = 14\): \(\sqrt{2\times14 + 28}-4\sqrt{14}=\sqrt{28 + 28}-4\sqrt{14}=\sqrt{56}-4\sqrt{14}=2\sqrt{14}-4\sqrt{14}=-2\sqrt{14}
eq0\). So the solution is \(n = 2\).

Answer:

A. \(n = 2\)