Problem 57 - Algebra for College Students

Question

Expressing solutions to the nearest one-thousandth. $$x^{2}+10 x-46=0$$

Step-by-Step Solution

Answer

The solutions are approximately 3.426 and -13.426.

Step 1 - Identify the formula

We will use the quadratic formula to solve for \(x\). The quadratic formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation \(ax^2 + bx + c = 0\).

Step 2 - Assign values to coefficients

Identify \(a\), \(b\), and \(c\) in the equation \(x^2 + 10x - 46 = 0\). Here, \(a = 1\), \(b = 10\), and \(c = -46\).

Step 3 - Substitute in the quadratic formula

Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula. We get: \[x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-46)}}{2 \cdot 1}\] Simplify under the square root: \[x = \frac{-10 \pm \sqrt{100 + 184}}{2}\] which becomes \[x = \frac{-10 \pm \sqrt{284}}{2}\].

Step 4 - Simplify the expression

Calculate \(\sqrt{284}\) then substitute it back into the equation: \(\sqrt{284} \approx 16.852\). So, the equation becomes \[x = \frac{-10 \pm 16.852}{2}\].

Step 5 - Solve for the roots

Calculate the two potential solutions for \(x\):1. \(x_1 = \frac{-10 + 16.852}{2} \approx \frac{6.852}{2} \approx 3.426\)2. \(x_2 = \frac{-10 - 16.852}{2} \approx \frac{-26.852}{2} \approx -13.426\)

Step 6 - Express the solutions

The solutions to the quadratic equation \(x^2 + 10x - 46 = 0\) are \(x \approx 3.426\) and \(x \approx -13.426\), rounded to the nearest one-thousandth.

Key Concepts

Quadratic Formula

The quadratic formula is a timeless tool in algebra, crucial for finding the roots of any quadratic equation. A quadratic equation is typically expressed in the standard form of \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are numeric coefficients where \( a eq 0 \). To find its solutions (or roots), we employ the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula involves three key parts:

• -b: This is the negation of the coefficient \( b \) and affects the direction of the parabola.
• \( \sqrt{b^2 - 4ac} \): Known as the discriminant, this expression determines the nature and number of roots. If it’s positive, we get two different real roots. If zero, exactly one real root. If negative, no real roots are present, implying two complex roots instead.
• \( \frac{1}{2a} \): This part essentially scales the root values.

The quadratic formula neatly simplifies finding solutions otherwise complicated by manual factorization or graphing.

Roots of Equations

Finding the roots of an equation means determining its solutions—values of \( x \) that make the equation true. For our quadratic equation \( x^2 + 10x - 46 = 0 \), these are the \( x \) values making the left-hand side equal zero. Roots can be understood through:

• Graphical Interpretation: The roots correspond to the points where the parabola \( y = x^2 + 10x - 46 \) intersects the x-axis.
• Number of Roots: As indicated by the quadratic formula's discriminant (\( b^2 - 4ac \)), a quadratic equation can have two, one, or no real roots.
• Nature of Roots:
  • Real:** Occur when the discriminant is zero or positive.
  • Complex:** Occur if the discriminant is negative, indicating non-real numbers.

In the case of our equation, substituting values into the quadratic formula provided two real roots: \( x \approx 3.426 \) and \( x \approx -13.426 \), which are computed by evaluating the quadratic formula's output.

Rounding Numbers

Rounding numbers is an essential skill, often used to simplify solutions or make them more understandable. When working with roots, especially from the quadratic formula, we frequently get lengthy decimal numbers. Here’s how to round to the nearest one-thousandth:

• Identify the place value: The third digit to the right of the decimal is the thousandths place.
• Check the fourth digit: If it's 5 or higher, increase the third digit by 1. If it's below 5, leave the third digit as is.

For example, rounding the number 3.4263 to the nearest thousandth gives us 3.426, because the digit in the ten-thousandths place (3) is less than 5. Similarly, \(-13.4264\) becomes \(-13.426\) when rounded to the nearest one-thousandth. This technique ensures clarity in solutions, providing a precise yet manageable result.