What is the solution of $\sqrt{x^2+49}=x+5$ ?
To find the solution to the equation \(\sqrt{x^2+49} = x+5\), we first identify the domain of the equation. Since the left-hand side is a square root, it must be greater than or equal to zero. Thus, for the equality to hold, the right-hand side must also be non-negative, giving us the condition \(x+5 \geq 0\), or \(x \geq -5\).
To eliminate the square root, we square both sides of the equation:
\[(\sqrt{x^2+49})^2 = (x+5)^2\]
Simplifying the left-hand side gives \(x^2 + 49\). Expanding the right-hand side, which is a binomial square, gives \(x^2 + 10x + 25\). Thus, we have:
\[x^2 + 49 = x^2 + 10x + 25\]
Subtracting \(x^2\) from both sides, the equation simplifies to a linear one:
\[49 = 10x + 25\]
Next, subtract 25 from both sides:
\[49 - 25 = 10x\]
\[24 = 10x\]
Solving for \(x\) by dividing by 10:
\[x = \frac{24}{10} = \frac{12}{5}\]
We check if \(x = \frac{12}{5}\) satisfies our initial condition \(x \geq -5\). Since \(\frac{12}{5} = 2.4 \geq -5\), the condition is met. Substituting \(x = 2.4\) back into the original equation:
Left-hand side: \(\sqrt{2.4^2 + 49} = \sqrt{5.76 + 49} = \sqrt{54.76} = 7.4\)
Right-hand side: \(2.4 + 5 = 7.4\)
Since both sides match, the solution is indeed \(x = \frac{12}{5}\).
This type of algebraic equation involving a square root appears in various physical contexts, such as calculating distances using the Pythagorean theorem where a coordinate is unknown. For instance, if you are tracking a moving object that maintains a certain distance from a fixed point (where the distance involves a square root), setting up and solving such an equation helps determine the exact position of the object at a given time. Mastering these techniques is fundamental for solving geometry and physics problems that require finding unknown variables related to spatial configurations.