Verified Answer :

  • Analyze the domain: the equation is defined for \(x \geq -1\).
  • Evaluate signs: the LHS (\(2\sqrt{x+1}\)) is always \(\geq 0\) and the RHS (\(-\sqrt{x+2}\)) is always \(\leq 0\).
  • Identify requirement: for equality, both sides must be \(0\).
  • Verify consistency: the equation \(x=-1\) for the LHS and \(x=-2\) for the RHS cannot be satisfied simultaneously, so there is no solution.
  • Conclusion: \(\boxed{\text{no solution}}\).

Explanation

  1. Domain Analysis

To solve the equation \(2 \sqrt{x+1} = -\sqrt{x+2}\), we first define the valid range for \(x\). The square root function \(\sqrt{u}\) is defined only for non-negative real numbers (\(u \geq 0\)). Therefore, we must satisfy two conditions:

  1. \(x+1 \geq 0 \implies x \geq -1\)
  2. \(x+2 \geq 0 \implies x \geq -2\)

The intersection of these conditions is \(x \geq -1\). This means any potential solution must be at least \(-1\).

  1. Range Analysis and Sign Comparison

Next, we examine the sign of each side of the equation for values of \(x\) in the domain \(x \geq -1\):

  1. The left-hand side (LHS) is \(2\sqrt{x+1}\). Since the principal square root is non-negative, \(2\sqrt{x+1} \geq 0\) for all \(x \geq -1\).
  2. The right-hand side (RHS) is \(-\sqrt{x+2}\). Since \(\sqrt{x+2} \geq 0\), its negative \(-\sqrt{x+2} \leq 0\) for all \(x \geq -1\).

For the equality \(2\sqrt{x+1} = -\sqrt{x+2}\) to hold, a non-negative number must equal a non-positive number. This can only occur if both sides are exactly equal to zero.

  1. Checking for Solutions and Conclusion

We now check if there is any value of \(x\) that makes both sides equal to zero:

  1. For the LHS to be zero: \(2\sqrt{x+1} = 0 \implies x+1 = 0 \implies x = -1\).
  2. For the RHS to be zero: \(-\sqrt{x+2} = 0 \implies x+2 = 0 \implies x = -2\).

Since the equation requires both sides to be zero, we need \(x = -1\) and \(x = -2\) to be true simultaneously. Because these two requirements are contradictory, no such \(x\) exists. Thus, the equation has no solution.

Examples

In many engineering and physics problems, equations involving square roots or other monotonic functions often appear. Identifying the domains and ranges of these functions can quickly reveal if a solution is physically possible without complex algebraic manipulation. For example, if you calculate that the power output of a system must be negative but your formula gives a positive square root expression, you know immediately that the model parameters must be adjusted. This approach is fundamental in ensuring that mathematical models accurately reflect real-world constraints.

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