Verified Answer :

  • Start with the original equation: \(\sqrt{x+10}-1=x\).
  • Isolate the radical term: \(\sqrt{x+10} = x+1\).
  • Square both sides to eliminate the radical: \(x+10 = (x+1)^2\).
  • Expand the squared binomial: \(x+10 = x^2+2x+1\).
  • The correct option is \(\boxed{x+10=x^2+2x+1}\).

Explanation

  1. Analysis of the Radical Equation

The objective is to determine which of the given options corresponds to the intermediate step obtained when solving the radical equation \(\sqrt{x+10}-1=x\). The standard approach to solving radical equations involves isolating the term containing the square root on one side and then squaring both sides to eliminate the radical.

  1. Isolating and Squaring the Radical

To solve the equation \(\sqrt{x+10}-1=x\), we first isolate the radical expression. By adding 1 to both sides of the equation, we obtain:
\[\sqrt{x+10} = x + 1\]
Next, to eliminate the square root, we square both sides of the equation:
\[(\sqrt{x+10})^2 = (x+1)^2\]
The left side simplifies to \(x+10\). Applying the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\) to the right side, we get:
\[(x+1)^2 = x^2 + 2x + 1\]
Thus, the equation after squaring both sides is \(x + 10 = x^2 + 2x + 1\).

  1. Comparing with Options and Conclusion

We compare our derived equation \(x + 10 = x^2 + 2x + 1\) with the provided options:

  • \(x+10=x^2+x+1\)
  • \(x+10=x^2+2x+1\)
  • \(x+10=x^2+1\)
  • \(x+10=x^2-1\)

The derived equation matches the second option. Therefore, the correct equation is \(\boxed{x+10=x^2+2x+1}\).

Examples

Understanding how to manipulate radical equations is essential in fields like physics and engineering, where equations often involve square roots from formulas such as the Pythagorean theorem, velocity in circular motion, or wave frequency calculations. For example, when calculating the distance between two points in a coordinate system or determining the time elapsed for a falling object, you might encounter radical equations that require careful algebraic manipulation to isolate and solve for the unknown variable, much like the process demonstrated here.