91Ó°ÊÓ

Quadratic equations play a crucial role in algebra, and they appear in various forms. A standard quadratic equation is expressed as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable.
In this specific problem, the equation

• appears in terms of the square root of \(x\)
• is initially given as \(x - 5\sqrt{x} + 4 = 0\)

To solve quadratic equations, the primary goal is to find the value(s) of \(x\) that satisfy the equation. Solving involves rewriting the equation into a familiar form, often requiring manipulation techniques such as factoring or using the quadratic formula, based on what makes the equation easiest to manage. While some quadratic equations can be solved simply by inspection, others may necessitate substitution or factoring to find the solution. Let’s explore these methods further below!

The substitution method simplifies solving quadratic equations by reducing complex expressions to a more manageable form. In our problem, we substitute \(y\) for \(\sqrt{x}\). This means:

• Let \(y = \sqrt{x}\)
• Then the equation \(x - 5\sqrt{x} + 4 = 0\) becomes \(y^2 - 5y + 4 = 0\)

This substitution transforms the equation into a straightforward quadratic in terms of \(y\). Through substitution, the operation of solving what seemed like a tricky expression is streamlined. Once you find the solutions for \(y\), resubstitute the values back into \(\sqrt{x}\) to find \(x\).
It’s worth noting that by substituting \(y = \sqrt{x}\), the problem becomes instantly more recognizable and manageable, allowing us to use other methods like factoring to find the equation's solutions efficiently.

Factoring is a widely-used technique for solving quadratic equations. By rewriting the equation as a product of binomials, it becomes easier to identify the solutions. The equation from the substitution method, \(y^2 - 5y + 4 = 0\), factors neatly into:

• \((y - 4)(y - 1) = 0\)

When an equation is written in this form, you can find potential solutions by setting each factor equal to zero:

• \(y - 4 = 0\) gives \(y = 4\)
• \(y - 1 = 0\) gives \(y = 1\)

These represent the values of \(y\) that solve the factored equation. The next and final step involves converting these \(y\) values back into the original terms of \(x\), using the resubstitution method we discussed. Thus, for\(\sqrt{x} = 4\) and \(\sqrt{x} = 1\), we square them to retrieve the original \(x\) values, leading to the solutions \(x = 16\) and \(x = 1\). Factoring makes solving complex equations more accessible by simplifying the problem into basic multiplicative steps.