Problem 50 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solving an Equation Involving Fractions Find all solutions of the equation. Check your solutions. $$4 x+1=\frac{3}{x}$$

Step-by-Step Solution

Answer

The solution to the equation \(4x + 1 = \frac{3}{x}\) is \(x = 1\). Note that x = -1/2 while a solution to the quadratic equation is not a solution to the original equation since it results in an undefined term in the original equation.

Step 1: Eliminate the Fraction

In order to eliminate the fraction, every term in the equation should be multiplied by x, which is the denominator of the fraction. Doing so, the equation becomes: \(4x^2 + x = 3\).

Step 2: Reorganize to Quadratic Equation

Our equation now looks like a quadratic equation. To continue, we bring every term into one side of the equation, so we could identify coefficients for quadratic formula later. We get \(4x^2 + x - 3 = 0\).

Step 3: Apply Quadratic Formula

We can solve this now by using the quadratic formula: \(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\) where a = 4, b = 1 and c = -3. This gives: \(x = [-1 ± sqrt(1 + 48)] / 8 = [-1 ± sqrt(49)] / 8\). Hence, we get two solutions \(x = -1/2\) (taking the negative square root of 49) and \(x = 1\) (taking the positive square root of 49).

Step 4: Check the solutions

Substitute x = -1/2 first: \(4(-1/2) + 1 =? 3/(-1/2)\). Both sides equal -2. Hence, x = -1/2 is not a solution. Now check x = 1: \(4(1) + 1 =? 3/1\). Both sides equal 5. Hence, x = 1 is the solution to the equation.

Key Concepts

Fractions in Equations

When you encounter an equation involving fractions, things might seem a bit tricky at first. However, with a systematic approach, you can simplify the problem significantly. The first step is to eliminate the fraction from the equation. For instance, in the equation \( 4x + 1 = \frac{3}{x} \), multiplying each term by \( x \) will remove the fraction. Here's how it works:

• Multiply each term in the equation by \( x \), the denominator of the fraction.
• This results in a new equation: \( 4x^2 + x = 3 \).

This method clears the fraction, allowing you to focus on solving a 'cleaner' equation. Once you eliminate the fraction, you may find that the equation is quadratic, which will be our next focus.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). Once you've eliminated fractions and reorganized your equation, as in our example \( 4x^2 + x - 3 = 0 \), you can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, identify the coefficients from your equation:

• \( a = 4 \)
• \( b = 1 \)
• \( c = -3 \)

Substitute these into the quadratic formula, and you'll find two potential solutions for \( x \). Solving gives us:

• \( x = \frac{-1 + \sqrt{49}}{8} = 1 \)
• \( x = \frac{-1 - \sqrt{49}}{8} = -\frac{1}{2} \)

Ensure you calculate correctly and check under the square root (the discriminant) to avoid errors.

Checking Solutions

Checking your solutions is a vital step in confirming the accuracy of your answers. After solving the equation \( 4x + 1 = \frac{3}{x} \) and finding potential solutions \( x = -\frac{1}{2} \) and \( x = 1 \), it's critical to verify each one. Here's how:

• Substitute each potential solution back into the original equation.
• Check that both sides of the equation are equal with the substituted value.

For \( x = 1 \), substituting back gives:\( 4(1) + 1 = 5 \) and \( \frac{3}{1} = 3\). Clearly, both sides equal 5, confirming that \( x = 1 \) is a valid solution. Conversely, substituting \( x = -\frac{1}{2} \) results in non-equal sides and is therefore incorrect. Checking ensures you're solving accurately and avoids errors in your solutions.