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4. solve: \\(\\sqrt{2x + 1} = 7\\)

Question

  1. solve: \\(\sqrt{2x + 1} = 7\\)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{2x + 1}=7\), we square both sides of the equation. Squaring the left side \((\sqrt{2x + 1})^2\) gives \(2x + 1\), and squaring the right side \(7^2\) gives \(49\). So we get the equation:
\(2x + 1 = 49\)

Step2: Solve for x

Subtract 1 from both sides of the equation \(2x+1 = 49\). Subtracting 1 from the left side gives \(2x\), and subtracting 1 from the right side gives \(49 - 1=48\). So we have:
\(2x=48\)
Then divide both sides by 2. Dividing the left side \(2x\) by 2 gives \(x\), and dividing the right side \(48\) by 2 gives \(24\). So:
\(x = 24\)

Step3: Check the solution (optional but recommended)

Substitute \(x = 24\) back into the original equation \(\sqrt{2x + 1}\). We have \(\sqrt{2\times24 + 1}=\sqrt{48 + 1}=\sqrt{49}=7\), which is equal to the right side of the original equation. So the solution is valid.

Answer:

\(x = 24\)