91Ó°ÊÓ

When solving any algebraic equation, isolating the variable is a critical first step. To do this, rearrange the equation so that the term containing the variable is on one side, while all other terms are on the opposite side.

In the case of solving quadratic equations, like the exercise \(5x^2 - 40 = 0\), you want to get \(x^2\) on its own. To achieve this, move the constant term \(40\) to the right-hand side by adding \(40\) to both sides, leading to \(5x^2 = 40\). This step creates a clearer pathway towards finding the value of the variable, \(x\).

The square root method is often the go-to technique for solving quadratics that can be written in the form \(x^2 = a\), where \(a\) is a positive real number. After isolating the variable, as shown previously, you obtain \(x^2 = 8\). Taking the square root of both sides is the next logical step: you apply the square root to \(x^2\) and also to \(8\).

This process will yield \(x = \pm\sqrt{8}\) because when you take the square root of a squared variable, you must consider both the positive and negative solutions.

Quadratics are polynomial equations of the second degree, generally written as \(ax^2 + bx + c = 0\). Our exercise simplifies to \(x^2 = a\) after isolating and simplifying the variable. To solve for \(x\), utilize the relationship between squares and square roots. Since squaring and taking the square root are inverse operations, applying the square root to both sides of \(x^2 = 8\) provides the values of \(x\) that satisfy the original equation: \(x = \pm\sqrt{8}\).

Remember that the solutions include both \(2\sqrt{2}\) and \(eg2\sqrt{2}\) because squaring either positive or negative \(2\sqrt{2}\) will result in \(8\). This information is crucial because it means there are usually two solutions to a quadratic equation when it's solvable by finding square roots.

Radical expressions involve roots, such as square roots, cube roots, etc. In the context of our exercise, once you've determined \(x = \pm\sqrt{8}\), it's often helpful to simplify the expression. Since \(8\) can be broken down into \(4\) times \(2\), and \(\sqrt{4}\) is a perfect square, you can simplify the expression as \(x = \pm 2\sqrt{2}\).

The simplified form makes it easier to understand and work with the value of \(x\). It's also more useful when substituting back into other equations or when graphing the function associated with the original equation.