91Ó°ÊÓ

A quadratic equation is an algebraic equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients with \(a eq 0\). The highest degree of the variable is 2, which is why it is called a quadratic equation. These equations can represent curves known as parabolas when graphed on a coordinate plane.

Quadratics can take various forms and not every quadratic equation needs to be complete with all three terms \(ax^2\), \(bx\), and \(c\). In the case of the problem given, only the \(ax^2\) term is present initially, simplifying the structure of the equation to \(5x^2 + 1 = 51\).

Quadratic equations have up to two solutions or roots. These solutions can be found using methods such as factoring, completing the square, or using the quadratic formula. However, in this particular instance, the equation involves a slightly adapted form known as the square root property.

Solving equations involves finding the value(s) of the variable that make the equation true. This means manipulating the equation until the variable is isolated on one side of the equation. Through systematic operations, the unknown becomes known.

The given equation \(5x^2 + 1 = 51\) is a quadratic equation that can be solved using the square root property. Solving begins with simplifying and transforming the equation:

• First, subtract 1 from both sides to get \(5x^2 = 50\).
• Next, divide every term by 5 to isolate \(x^2\), which gives \(x^2 = 10\).

These steps prepare the equation for the application of the square root property to find \(x\).

This process of solving entails performing operations that maintain the balance of the equation while simplifying it.

In algebra, calculations involve using arithmetic operations in combination with algebraic principles to solve equations or simplify expressions.

In the context of the square root property, after isolating \(x^2\), the algebra calculation involves applying the square root to both sides of the equation. Mathematically, this step is expressed as:

• Taking the square root of \(x^2 = 10\), we derive \(x = \pm\sqrt{10}\).

Remember that every positive real number has two square roots: one positive and one negative. Hence, it's crucial to consider both \(+\) and \(-\) solutions when dealing with quadratic equations.

This demonstrates the elegance of algebra calculations, which involve systematically applying rules and properties to uncover the desired solutions.