Problem 29 - Elementary Algebra

Question

Solve. $$ x_{2}-14 x+40=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 10\) and \(x = 4\).

Step 1: Identify the equation type

The given equation is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -14\), and \(c = 40\).

Step 2: Use the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). To solve for \(x\), we'll substitute \(a = 1\), \(b = -14\), and \(c = 40\).

Step 3: Calculate the discriminant

The discriminant \(D\) is calculated as \(b^2 - 4ac\). Substitute the values to find \(D = (-14)^2 - 4 \times 1 \times 40\).

Step 4: Perform the discriminant calculation

Calculate: \((-14)^2 = 196\) and \(4 \times 1 \times 40 = 160\), so \(D = 196 - 160 = 36\).

Step 5: Solve for x using the quadratic formula

Since \(D = 36\), which is positive, the equation has two real roots. Substitute into the quadratic formula:\[x = \frac{-(-14) \pm \sqrt{36}}{2 \times 1} = \frac{14 \pm 6}{2}\]

Step 6: Calculate the roots

Calculate the roots using \(14 \pm 6\) as follows: 1. For \(14 + 6\): \[x = \frac{14 + 6}{2} = \frac{20}{2} = 10\]2. For \(14 - 6\): \[x = \frac{14 - 6}{2} = \frac{8}{2} = 4\]

Step 7: State the solutions

The solutions for the equation \(x^2 - 14x + 40 = 0\) are \(x = 10\) and \(x = 4\).

Key Concepts

Quadratic Formula

Understanding the quadratic formula is key to solving quadratic equations. It provides a simple way to find the roots of any quadratic equation, regardless of whether it can be factored easily. The quadratic formula is: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
This formula is derived from the standard form of a quadratic equation \(ax^2 + bx + c = 0\).
Here’s how it works:

• \(a\), \(b\), and \(c\) are coefficients from the quadratic equation.
• You first calculate the discriminant, \(b^2 - 4ac\), which is the expression under the square root.
• The symbol \(\pm\) indicates that there are usually two solutions.

Using the quadratic formula involves plugging these values into the formula and solving for \(x\). It’s particularly useful for equations that don’t factor neatly.

Discriminant

The discriminant is a powerful concept that tells you a lot about the nature of the roots of a quadratic equation. It is given by the expression \(b^2 - 4ac\).

• If the discriminant is positive, there are two distinct real roots, meaning the parabola crosses the x-axis at two points.
• If the discriminant is zero, there is exactly one real root, indicating the parabola just touches the x-axis.
• If the discriminant is negative, there are no real roots, and the solutions are complex or imaginary.

In our example, the discriminant was calculated as \(36\), which is positive. Hence, the equation has two real roots, and the quadratic intercepts the x-axis at two separate points.

Real Roots

Real roots are the solutions to quadratic equations where the discriminant is zero or positive. They’re called "real" because they are real numbers, as opposed to imaginary numbers.
Real roots can be:

• Distinct: When the discriminant is positive, yielding two different solutions.
• Repeated: When the discriminant is zero, and both solutions are the same.

In the given problem, since the discriminant is \(36\), which is greater than zero, we find two distinct real roots: \(x = 10\) and \(x = 4\). These roots represent the x-values where the quadratic equation intersects the x-axis.

Factoring Quadratics

Factoring quadratics is another method to find the roots and is known for its straightforwardness with simpler equations. It involves rewriting the quadratic equation in a product of binomials form, such as \((x - r)(x - s) = 0\).
Here's a quick look at the steps:

• Write the equation in standard form.
• Find two numbers that multiply to give \(c\) and add up to \(b\).
• Rewrite the middle term using these two numbers and factor by grouping.

For our quadratic, which is \(x^2 - 14x + 40 = 0\), we can factor it as \((x - 10)(x - 4) = 0\).
This gives the solutions directly as the roots \(x = 10\) and \(x = 4\), confirming the solution we found using the quadratic formula.