- Simplify the equation using \(\sqrt{(x^2-3)^2} = |x^2-3|\), yielding \(|x^2-3| = 2x+12\).
- Analyze the condition \(x \geq -6\) for the right-hand side.
- Solve the two cases for the absolute value: \(x^2-3 = 2x+12\) and \(3-x^2 = 2x+12\).
- Verify the solutions \(x=5\) and \(x=-3\), which are both valid.
- The final solution is \(\boxed{x=-3, x=5}\).
Explanation
- Simplifying the Equation
We are given the equation \(\sqrt{x^4-6x^2+9} = 2(x+6)\).
First, we analyze the expression under the square root: \(x^4-6x^2+9 = (x^2-3)^2\).
Thus, the equation can be written as \(\sqrt{(x^2-3)^2} = 2(x+6)\).
Recalling that \(\sqrt{y^2} = |y|\) for any real number \(y\), the equation simplifies to \(|x^2-3| = 2(x+6)\).
Because the left-hand side is an absolute value, it is non-negative, meaning the right-hand side must also be non-negative: \(2(x+6) \geq 0\), which implies \(x \geq -6\).
- Case 1: The expression inside the absolute value is non-negative
To solve \(|x^2-3| = 2(x+6)\), we split the problem into two cases based on the sign of the expression inside the absolute value:
Case 1: \(x^2 - 3 \geq 0\)
This condition holds when \(x \geq \sqrt{3}\) or \(x \leq -\sqrt{3}\). In this case, \(|x^2-3| = x^2-3\).
The equation becomes:
\(x^2 - 3 = 2x + 12\)
\(x^2 - 2x - 15 = 0\)
We factor this quadratic equation:
\((x-5)(x+3) = 0\)
This gives us potential solutions \(x=5\) and \(x=-3\).
Checking these against the conditions \(x \geq \sqrt{3}\) or \(x \leq -\sqrt{3}\) and \(x \geq -6\):
- For \(x=5\): \(5^2 = 25 \geq 3\) and \(5 \geq -6\). This solution is valid.
- For \(x=-3\): \((-3)^2 = 9 \geq 3\) and \(-3 \geq -6\). This solution is valid.
- Case 2: The expression inside the absolute value is negative
Case 2: \(x^2 - 3 < 0\)
This condition holds when \(-\sqrt{3} < x < \sqrt{3}\). In this case, \(|x^2-3| = -(x^2-3) = -x^2+3\).
The equation becomes:
\(-x^2 + 3 = 2x + 12\)
\(x^2 + 2x + 9 = 0\)
To find the roots, we check the discriminant \(\Delta = b^2 - 4ac\):
\(\Delta = 2^2 - 4(1)(9) = 4 - 36 = -32\)
Since the discriminant is negative, there are no real solutions for this case.
- Conclusion
Combining the results from the two cases, the only valid solutions are \(x=5\) and \(x=-3\). Therefore, the solution set for the given equation is \(x=-3, x=5\).
Examples
Absolute value equations are frequently used in physics and engineering to model systems with tolerances or boundaries. For example, if you are manufacturing a part that needs to be within a certain distance from a target dimension (like a sensor that must be within a specific range from a center point), the absolute value function helps represent the error margin. Solving such equations allows you to determine the precise range of parameters that satisfy these design constraints.