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Solving quadratic equations is a fundamental skill in algebra. These equations typically have the form \(ax^2 + bx + c = 0\). However, when dealing with equations like \(z^2 + 5 = 19\), we can simplify our approach using the square root property. To start, aim to express the equation in a format where one side is a perfect square. Doing this paves the way for using the square root method effectively. In some cases, like the one we have, the equation \(z^2 + 5 = 19\) simplifies to \(z^2 = 14\) once any constants are eliminated, letting us proceed efficiently with our solution. Using methods like completing the square or the quadratic formula can solve more complex quadratics, but for simple ones as here, isolating the squared term and taking square roots can be much quicker.

One crucial step in solving quadratic equations via the square root property is isolating the squared term on one side of the equation. Here’s how you can do it:

• Start with your quadratic equation. In our case: \(z^2 + 5 = 19\).
• Subtract the constant term from both sides to "move it away" from the squared term. For our example, subtract 5 from both sides, giving us: \(z^2 = 19 - 5\).
• Simplify the right side of the equation: \(z^2 = 14\).

By isolating \(z^2\), we prepare the equation for the application of the square root property. Keeping the squared term alone on one side is key to simplifying the equation to a basic square form. This step makes taking the square root straightforward.

When solving quadratic equations using the square root property, it's essential to remember that taking the square root of a number yields both positive and negative solutions. Here's why this happens:

• The square of both positive and negative numbers is positive. For example, \((3)^2\) and \((-3)^2\) both result in 9.
• Therefore, when solving \(z^2 = 14\), applying the square root property we write: \(z = \pm \sqrt{14}\).

This notation \(\pm\) (read as "plus or minus") indicates there are two solutions: one positive \(\sqrt{14}\) and one negative \(-\sqrt{14}\). In many algebra problems, accounting for these dual solutions is vital because quadratic equations derive from symmetrical graphs (parabolas) that can intersect the x-axis at two points. Understanding this inherent nature of quadratic solutions prevents overlooking possible roots, ensuring a complete and accurate solution.