Problem 119 - College Algebra Essentials

Question

Describe two methods for solving this equation: \(x-5 \sqrt{x}+4=0\)

Step-by-Step Solution

Answer

By factoring or using the quadratic formula, we find that the solutions to \(x - 5\sqrt{x} + 4 = 0\) are \(x = 1\) and \(x = 16\).

Step 1: Method 1: Factoring

First, rewrite the equation as \((\sqrt{x})^2 - r\sqrt{x} + s = 0\), where \(r=5\) and \(s=4\). By factoring, we might rewrite this as \((\sqrt{x} - a)(\sqrt{x} - b) = 0\), where \(a\) and \(b\) are factors of 4 that add up to 5. This can be done by realizing that 4 and 1 are the proper numbers. So, \((\sqrt{x} - 4)(\sqrt{x} - 1) = 0\). Setting these factors to zero individually gives us possible solutions for \(x\): \(\sqrt{x} = 4\) and \(\sqrt{x} = 1\), which further gives \(x = 16\) and \(x = 1\).

Step 2: Method 2: Quadratic Formula

Another way to solve the question would be to implement the quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this equation, \(a\) corresponds to the coefficient of the term \((\sqrt{x})^2\), \(b\) denotes the coefficient of the term \(\sqrt{x}\), and \(c\) is the constant term. Here, \(a = 1\), \(b = -5\), and \(c = 4\). Substituting these into the formula, we get: \(\frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)} = \frac{5 \pm \sqrt{25 - 16}}{2}\), which simplifies to two possible solutions for \(\sqrt{x}\): \(\sqrt{x} = 4\) and \(\sqrt{x} = 1\), giving \(x = 16\) and \(x = 1\).