Problem 21 - Elementary Algebra

Question

Solve. $$x 2=14 x+3$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 7 + 2\sqrt{13}\) and \(x = 7 - 2\sqrt{13}\).

Step 1: Simplify the Equation

First, look at the given equation: \(x^2 = 14x + 3\). Notice that it is a quadratic equation. To solve it, we need to move all terms to one side of the equation so that one side equals zero. To accomplish that, subtract \(14x\) and \(3\) from both sides: \(x^2 - 14x - 3 = 0\).

Step 2: Identify the Coefficients

With the equation \(x^2 - 14x - 3 = 0\) now in standard quadratic form \(ax^2 + bx + c = 0\), we identify the coefficients: \(a = 1\), \(b = -14\), and \(c = -3\).

Step 3: Use the Quadratic Formula

Recall the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the coefficients \(a = 1\), \(b = -14\), and \(c = -3\) into the formula.

Step 4: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\) where \(b = -14\), \(a = 1\), and \(c = -3\). The expression becomes \((-14)^2 - 4(1)(-3) = 196 + 12 = 208\).

Step 5: Solve for x

Now substitute back into the quadratic formula: \(x = \frac{-(-14) \pm \sqrt{208}}{2(1)}\), simplifying to \(x = \frac{14 \pm \sqrt{208}}{2}\). Calculate: \(\sqrt{208} = 4\sqrt{13}\). So, \(x = \frac{14 \pm 4\sqrt{13}}{2}\).

Step 6: Simplify the Solutions

Break down the expression \(x = \frac{14 \pm 4\sqrt{13}}{2}\) into simpler terms: \(x = \frac{14}{2} \pm \frac{4\sqrt{13}}{2}\), which simplifies to \(x = 7 \pm 2\sqrt{13}\).

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations, which are equations in the form of \(ax^2 + bx + c = 0\). This formula provides a method to find the roots of any quadratic equation, regardless of whether it can be factored easily or not.
It is expressed as:

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

Here,

• \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation.
• The symbol \(\pm\) indicates that there are generally two possible solutions for \(x\).

Using this formula, you can directly substitute the values of \(a\), \(b\), and \(c\) and find the solutions for \(x\). It's particularly valuable when factoring is complex or when the roots are irrational numbers. Understanding how to apply this formula is fundamental in algebra.

Discriminant

The discriminant is a part of the quadratic formula that gives us information about the nature of the roots of the quadratic equation. It is the expression under the square root in the quadratic formula: \(b^2 - 4ac\).
The value of the discriminant can tell us a lot:

• If \(b^2 - 4ac > 0\), there are two distinct real roots. This means the graph of the quadratic equation will intersect the x-axis at two points.
• If \(b^2 - 4ac = 0\), there is exactly one real root, or a repeated root. This means the graph touches the x-axis at one point, known as a vertex.
• If \(b^2 - 4ac < 0\), there are no real roots. Instead, the solutions are complex or imaginary numbers, and the graph does not intersect the x-axis at all.

In the exercise, the discriminant \(b^2 - 4ac\) was calculated as 208, which is greater than zero, indicating that our quadratic equation has two distinct real roots.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of \(x\) that satisfy the equation. When you have a quadratic equation like \(x^2 - 14x - 3 = 0\), the quadratic formula is often the go-to method for finding the solutions.
Here's a step-by-step overview for using the quadratic formula:

• Ensure your quadratic equation is in standard form \(ax^2 + bx + c = 0\).
• Identify your coefficients \(a\), \(b\), and \(c\).
• Substitute these values into the quadratic formula.
• Calculate the discriminant \(b^2 - 4ac\).
• Use the value of the discriminant to understand the nature of the roots.
• Finally, solve for \(x\) using the formula and simplify if needed.

By substituting \(a = 1\), \(b = -14\), and \(c = -3\) into the formula, you evaluated the discriminant and then solved for \(x\). The result \(x = 7 \pm 2\sqrt{13}\) provides the two possible solutions for \(x\), showcasing how effective the quadratic formula is for handling such problems.