91Ó°ÊÓ

Quadratic equations form the basis of a wide variety of problems in mathematics. They often appear in the form \( ax^2 + bx + c = 0 \). In our case, the quadratic equation is \( 2x^2 = 250 \). Quadratic equations can be solved using different methods such as factoring, completing the square, using the quadratic formula, or taking square roots; the latter is often used for equations without a linear component \(b = 0\).
Our first step in solving the equation \(2x^2 = 250\) involved dividing both sides of the equation by 2. This simplifies the equation to \(x^2 = 125\), isolating the \(x^2\) term, which is crucial for the next step.
Once the equation is simplified to \(x^2 = n\), taking the square root of both sides allows us to solve for \(x\). This step produces two solutions: one positive and one negative, since squaring either a positive or negative number results in the same product.

Square roots are fundamental to the world of mathematics. The square root of a number \( n \) is a value that, when multiplied by itself, equals \( n \). The square root operation is denoted using the radical symbol \( \sqrt{ } \). For example, \( \sqrt{4} = 2 \) because \(2 \times 2 = 4\).
In the exercise, once we had the simplified form \(x^2 = 125\), we took the square root of both sides to get \(x = \sqrt{125}\) and \(x = -\sqrt{125}\). Remember: \(x^2\) could mean \(x\) is both positive and negative when under a square root. This is why solutions include both \( \pm \sqrt{n} \). This concept helps solve quadratic equations directly when a term is already isolated as a square.

Simplifying square roots is a powerful skill that makes equations easier to interpret and solve. To simplify \(\sqrt{125}\), we first decompose 125 into its prime factors. Note that 125 can be expressed as \(25 \times 5\). Since \( 25 \) is a perfect square (because \( 5 \times 5 = 25 \)), we can simplify \( \sqrt{125} \) into \( \sqrt{25 \times 5} \).

This simplifies further, separating the terms to \( \sqrt{25} \times \sqrt{5} \). We know that \( \sqrt{25} = 5 \), giving us \( 5\sqrt{5} \). Therefore, the solutions to the equation become \(x = 5\sqrt{5}\) and \(x = -5\sqrt{5}\).
This process highlights the importance of recognizing and extracting perfect squares within the radicand – the number inside the square root symbol.