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Square roots are everyday math operations that help find the original value of a number when it is multiplied by itself. In the context of solving quadratic equations, square roots allow us to "undo" the square of a variable. Let's dive deeper into how this works.

Whenever you see an expression like \((x-3)^2 = 3\), the square root operation helps transform the equation to \(x-3 = \pm\sqrt{3}\). Note here the use of the "\(\pm\)" symbol. Since squaring a number always gives a positive result, taking the square root returns two possible values: one positive and one negative.

In practice:

• The square root of 4 is 2, because \(2^2 = 4\), and also -2 because \((-2)^2 = 4\).
• Thus for \((x-3)^{2} = 3\), \(x-3\) could be either \(\sqrt{3}\) or \(-\sqrt{3}\).

That's why equations involving squares often have two solutions.

"Isolating the variable" is a critical concept when solving equations. It means adjusting the equation to get the unknown variable, usually represented as \(x\), on one side of the equation all by itself. This is an important step because it tells us the value of \(x\) directly.

In our example equation \((x-3) = \pm\sqrt{3}\), we've already taken the square root to trim down the equation. Now, we need to make \(x\) stand alone by eliminating any numbers around it. Here's how we isolate \(x\):

• Add 3 to both sides of the equation \((x-3) = \pm\sqrt{3}\) to remove the \(-3\) and get \(x = 3 \pm \sqrt{3}\).
• This tells us that you have two equations: \(x = 3 + \sqrt{3}\) and \(x = 3 - \sqrt{3}\).

By performing these operations step by step, you systematically peel away each element influencing the variable, until you discover the solutions for \(x\).

Quadratic equations are vital tools in algebra, used to find solutions for equations of the form \(ax^2 + bx + c = 0\). There are several methods to solve these equations, two of which involve using square roots and isolating variables, as demonstrated in our example.

When solving \(7(x-3)^2 = 21\), first simplify the equation by dividing both sides by 7. This results in \((x-3)^2 = 3\).

Next, employ square roots to eliminate the squared term, leading to \(x-3 = \pm\sqrt{3}\). By isolating \(x\), we get \(x = 3\pm\sqrt{3}\). Each \(\pm\) allows us to derive two potential answers:

• Increasing by the square root gives \(x = 3 + \sqrt{3}\).
• Decreasing by the square root gives \(x = 3 - \sqrt{3}\).

These values, \(3 + \sqrt{3}\) and \(3 - \sqrt{3}\), are solutions to the quadratic equation. This approach highlights the symmetry of quadratic equations, as they often yield two solutions that are equidistant from a central point on the number line.