Problem 10 - Beginning and Intermediate Algebra

Question

Solve each equation. Check the solutions. \(4-\frac{7}{r}-\frac{2}{r^{2}}=0\)

Step-by-Step Solution

Answer

The solutions are \( r = 2 \) and \( r = -\frac{1}{4} \).

Step 1: Set up the equation

Start by rewriting the given equation: \[ 4 - \frac{7}{r} - \frac{2}{r^2} = 0 \]

Step 2: Multiply through by the least common denominator

Multiply every term by \( r^2 \) to clear the denominators: \[ 4r^2 - 7r - 2 = 0 \]

Step 3: Solve the quadratic equation

Solve the quadratic equation \( 4r^2 - 7r - 2 = 0 \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -7 \), and \( c = -2 \). First, calculate the discriminant: \[ b^2 - 4ac = (-7)^2 - 4(4)(-2) = 49 + 32 = 81 \] Then find the roots: \[ r = \frac{7 \pm \sqrt{81}}{8} \] \[ r = \frac{7 \pm 9}{8} \] So, the solutions are: \[ r_1 = \frac{16}{8} = 2 \] \[ r_2 = \frac{-2}{8} = -\frac{1}{4} \]

Step 4: Check the solutions

Substitute each solution back into the original equation to verify: For \( r = 2 \): \[ 4 - \frac{7}{2} - \frac{2}{2^2} = 4 - 3.5 - 0.5 = 0 \] For \( r = -\frac{1}{4} \): \[ 4 - \frac{7}{-\frac{1}{4}} - \frac{2}{(-\frac{1}{4})^2} = 4 + 28 - 32 = 0 \] Both solutions satisfy the original equation.

Key Concepts

Solving Quadratic Equations

Quadratic equations are mathematical expressions of the form \[ax^2 + bx + c = 0\]. Solving these equations involves finding the values of the variable, usually represented by \(x\), that make the equation true. A quadratic equation is characterized by its highest exponent being 2, which means it includes a square term. In the given problem, we transformed the original equation by multiplying through by the least common denominator, \(r^2\), to remove fractions. This gave us a more familiar form that we can solve for \(r\): \[4r^2 - 7r - 2 = 0\].

Quadratic Formula

The quadratic formula is a universal tool for solving any quadratic equation. It is given by \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Here, \(a,\) \(b,\), and \(c\) are the coefficients from the standard quadratic equation \(ax^2 + bx + c = 0\). In our exercise, we identified \(a = 4\), \(b = -7\), and \(c = -2\). After substituting these values into the quadratic formula, we obtain the possible values of \(r\). The result offers two potential solutions due to the \(\pm\) sign in the formula, which represents the two roots of the quadratic equation.

Discriminant

The discriminant helps us determine the nature and number of solutions of a quadratic equation. It is found inside the quadratic formula and is given by \[b^2 - 4ac\]. The value of the discriminant can tell us whether we have two real solutions, one real solution, or no real solutions. In our problem, we calculated the discriminant as follows: \((-7)^2 - 4(4)(-2) = 49 + 32 = 81\). A positive discriminant (like 81 here) indicates two distinct real solutions. This is why we got two roots: \(r = 2\) and \(r = -\frac{1}{4}\).

Checking Solutions

It’s important to verify that the solutions obtained are correct by substituting them back into the original equation. This ensures that no mathematical errors were made during the solving process. For \(r = 2\), substituting into the original equation gives:\[4 - \frac{7}{2} - \frac{2}{2^2} = 0\]. For \(r = -\frac{1}{4}\), substituting back gives:\[4 + 28 - 32 = 0\]. Both checks confirm that our solutions satisfy the original equation. This final verification step is crucial for ensuring the accuracy of your solution.