Problem 43 - Intermediate Algebra

Question

Solve. \(\sqrt{5 x-1}-\sqrt{x}+2=3\)

Step-by-Step Solution

Answer

The solution is \(x = 1\).

Step 1: Isolate the Radical

First, subtract 2 from both sides of the equation to begin isolating the radical term. \[ \sqrt{5x-1} - \sqrt{x} = 1 \]

Step 2: Eliminate the Second Radical

Isolate the first radical by adding \(\sqrt{x}\) to both sides of the equation:\[ \sqrt{5x-1} = 1 + \sqrt{x} \]

Step 3: Square Both Sides

Square both sides to eliminate the square root:\[ (\sqrt{5x-1})^2 = (1 + \sqrt{x})^2 \]Which simplifies to \[ 5x - 1 = 1 + 2\sqrt{x} + x \]

Step 4: Rearrange the Equation

Move all terms except the remaining radical to one side of the equation:\[ 4x - 2 = 2\sqrt{x} \]

Step 5: Isolate Square Root Again

Divide every term by 2 to simplify the equation:\[ 2x - 1 = \sqrt{x} \]

Step 6: Square Again

Square both sides once more to eliminate the square root:\[ (2x - 1)^2 = x \]Expanding gives \[ 4x^2 - 4x + 1 = x \]

Step 7: Form a Quadratic Equation

Rearrange the terms to form a quadratic equation:\[ 4x^2 - 5x + 1 = 0 \]

Step 8: Solve the Quadratic Equation

Use the quadratic formula to solve for \(x\):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our equation, \(a = 4\), \(b = -5\), and \(c = 1\), thus\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]\[ x = \frac{5 \pm \sqrt{25 - 16}}{8} \]\[ x = \frac{5 \pm \sqrt{9}}{8} \]\[ x = \frac{5 \pm 3}{8} \]This results in two potential solutions: \(x = 1\) and \(x = \frac{1}{4}\).

Step 9: Verify Each Solution

Verify each potential solution by substituting back into the original equation to ensure it holds true.For \(x = 1\):\[ \sqrt{5(1)-1} - \sqrt{1} + 2 = 3 \Rightarrow \sqrt{4} - 1 + 2 = 3 \Rightarrow 3 = 3 \]This solution checks out.For \(x = \frac{1}{4}\):\[ \sqrt{5(\frac{1}{4})-1} - \sqrt{\frac{1}{4}} + 2 = 3 \Rightarrow \sqrt{\frac{5}{4} - 1} - \frac{1}{2} + 2 = 3 \Rightarrow \sqrt{\frac{1}{4}} - \frac{1}{2} + 2 = 3 \Rightarrow \frac{1}{2} - \frac{1}{2} + 2 eq 3 \]This solution does not satisfy the equation.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial expressions of the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Solving quadratic equations often involves finding the values of \(x\) that satisfy the equation. These roots can be real or complex numbers, depending on the discriminant \(b^2 - 4ac\). This is the part under the square root in the quadratic formula:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root, known as a repeated root.
• If the discriminant is negative, the equation has two complex roots.

In our solved example, after isolating and squaring the radicals, we reached a quadratic equation: \(4x^2 - 5x + 1 = 0\). By applying the quadratic formula, you can determine that the potential solutions are \(x = 1\) and \(x = \frac{1}{4}\).

Radicals

Radicals, particularly square roots, appear frequently in various mathematical problems. A radical involves an expression under a root sign, like \(\sqrt{x}\). When solving equations with radicals, it is important to isolate the radical before eliminating it through squaring. This prevents incorrect cancellations or calculations.When you square a radical, such as \(\sqrt{y}\), the radical sign disappears, turning it into a regular algebraic term \(y\). In the task we tackled, radicals were addressed twice. First, the initial equation \(\sqrt{5x-1} - \sqrt{x} = 1\) was manipulated to \(\sqrt{5x-1} = 1 + \sqrt{x}\), and then squared to eliminate the square roots. However, energetic caution is required when dealing with radicals. Squaring can introduce extraneous solutions—values that solve the squared equation but not the original. Hence, it's crucial to verify solutions in the original equation.

Verification of Solutions

Verification ensures that solutions found actually satisfy the original equation. With the presence of radicals in equations, it's not uncommon to produce extraneous solutions while solving. Verification is done by substituting the value of \(x\) back into the original equation to check if it produces a true statement.In the current example, after calculating solutions \(x = 1\) and \(x = \frac{1}{4}\) from the quadratic equation, verification showed only \(x = 1\) holds true in \[\sqrt{5 \times 1 - 1} - \sqrt{1} + 2 = 3\] The verification step rejected \(x = \frac{1}{4}\), as \[\sqrt{5 \times \frac{1}{4} - 1} - \sqrt{\frac{1}{4}} + 2 eq 3\] This thorough check ensures that the final solution is truly valid, preserving the robustness of mathematical problem-solving.