Problem 66 - Intermediate Algebra

Question

The solution set for \(x^{2}-4 x-37=0\) is \(\\{2 \pm \sqrt{41}\\}\). With a calculator, we found a rational approximation, to the nearest one-thousandth, for each of these solutions. $$ 2-\sqrt{41}=-4.403 \quad \text { and } \quad 2+\sqrt{41}=8.403 $$ Thus the solution set is \(\\{-4.403,8.403\\}\), with the answers rounded to the nearest one-thousandth. Solve each of the equations in Problems 64-73, expressing solutions to the nearest one-thousandth. $$ x^{2}+6 x-44=0 $$

Step-by-Step Solution

Answer

The solutions are approximately 4.280 and -10.280.

Step 1: Identify the Equation

We are given the quadratic equation \(x^2 + 6x - 44 = 0\). Our task is to find the solutions to this equation and express them to the nearest thousandth.

Step 2: Calculate the Discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). Here, \(a=1\), \(b=6\), and \(c=-44\). Calculate the discriminant: \(D = 6^2 - 4(1)(-44) = 36 + 176 = 212\).

Step 3: Apply the Quadratic Formula

The solutions of the quadratic equation \(ax^2 + bx + c = 0\) are given by the formula \(x = \frac{-b \pm \sqrt{D}}{2a}\). Substitute \(a=1\), \(b=6\), and \(D=212\) into the formula: \[x = \frac{-6 \pm \sqrt{212}}{2}\].

Step 4: Calculate the Square Root

Compute \(\sqrt{212}\) using a calculator. \(\sqrt{212} \approx 14.560\) to the nearest thousandth.

Step 5: Find Both Solutions

Substitute \(\sqrt{212} \approx 14.560\) into the quadratic formula to find both solutions. For the first solution: \(x_1 = \frac{-6 + 14.560}{2} \approx 4.280\). For the second solution: \(x_2 = \frac{-6 - 14.560}{2} \approx -10.280\).

Step 6: Round the Solutions

Round both solutions to the nearest thousandth. The solutions are \(x_1 \approx 4.280\) and \(x_2 \approx -10.280\).

Key Concepts

Quadratic Formula

The quadratic formula is a handy tool for solving quadratic equations like the one we're working with here, which is of the form \( ax^2 + bx + c = 0 \). It's particularly useful when the equation doesn't easily factorize. The formula itself is expressed as: \[ x = \frac{-b \pm \sqrt{D}}{2a}\] where \( D \) is the discriminant, a key part of the solution that we'll explore further. In this case, the coefficients are \( a = 1 \), \( b = 6 \), and \( c = -44 \). Inserting these values into the formula allows us to solve for \( x \):

• Calculate \( \frac{-b}{2a} \), which provides the central axis of the parabola represented by the quadratic equation. For our equation, this part calculates to \( \frac{-6}{2(1)} = -3 \).
• The "\( \pm \sqrt{D} \)" part means you'll end up with two potential solutions, rooted in the nature of quadratic equations having two roots.

Using the quadratic formula is especially essential when the discriminant is a non-perfect square, which we'll get into next.

Discriminant Calculation

The discriminant \( D \) is a component of the quadratic formula and it helps determine the nature of the roots of the quadratic equation. It's calculated using the formula: \[ D = b^2 - 4ac\] Let's plug the values from our equation \( x^2 + 6x - 44 = 0 \):

• \( a = 1 \)
• \( b = 6 \)
• \( c = -44 \)

Substituting these into the formula gives us: \[ D = 6^2 - 4 \times 1 \times (-44) = 36 + 176 = 212\] The discriminant value of \( 212 \) is positive, signifying that our quadratic equation has two distinct real roots. The fact that \( D \) is not a perfect square also suggests that these roots are irrational, meaning they can't be expressed as simple fractions, and will have a more complex decimal expansion.

Rounding Solutions

Once we've used the quadratic formula, the final step is to round the solutions to the appropriate decimal place for usability and clarity. When we calculate the square root of the discriminant in our example: \[ \sqrt{212} \approx 14.560\]we get a decimal approximation that needs rounding. The problem asks for solutions rounded to the nearest thousandth, which involves:

• For \( x_1 \, (\frac{-6 + 14.560}{2}) \), we find the result to be approximately \( 4.280 \).
• For \( x_2 \, (\frac{-6 - 14.560}{2}) \), we arrive at approximately \( -10.280 \).

Rounding ensures that we express these solutions in a practical and understandable way, especially when exact numbers aren't necessary or possible. It's crucial to follow the standard rounding rules: if the digit at the thousandth place is followed by 5 or more, increase the thousandth digit by one, ensuring precision.