Problem 26 - Algebra and Trigonometry

Question

Find all real solutions of the equation. \(x^{2}+5 x-6=0\)

Step-by-Step Solution

Answer

The solutions are \(x = 1\) and \(x = -6\).

Step 1: Identify the equation type

The equation given is a quadratic equation of the form \(ax^2 + bx + c = 0\). In this case, \(a = 1\), \(b = 5\), and \(c = -6\).

Step 2: Use the quadratic formula

The quadratic formula to find the roots of \(ax^2 + bx + c = 0\) is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). We will use this formula to find the solutions for \(x^2 + 5x - 6 = 0\).

Step 3: Calculate the discriminant

The discriminant, \(D\), is given by \(b^2 - 4ac\). Calculate \(D\): \(D = 5^2 - 4 \cdot 1 \cdot (-6) = 25 + 24 = 49\).

Step 4: Find the square root of the discriminant

Since \(D = 49\), we find \(\sqrt{D} = \sqrt{49} = 7\).

Step 5: Solve for the roots

Substitute \(b = 5\), \(\sqrt{D} = 7\), \(a = 1\) into the quadratic formula: \(x = \frac{-5 \pm 7}{2 \cdot 1}\). This gives two possible solutions for \(x\): 1) \(x_1 = \frac{-5 + 7}{2} = \frac{2}{2} = 1\) 2) \(x_2 = \frac{-5 - 7}{2} = \frac{-12}{2} = -6\).

Step 6: Verify the solutions

To ensure the solutions are correct, substitute \(x = 1\) and \(x = -6\) back into the original equation \(x^2 + 5x - 6 = 0\). For \(x = 1\): \((1)^2 + 5(1) - 6 = 1 + 5 - 6 = 0\) (True). For \(x = -6\): \((-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0\) (True). Thus, both solutions are verified.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). A quadratic equation features a variable squared (such as \( x^2 \)), and turns up quite often in algebra. The quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]allows you to find the solutions (or roots) of these equations. Here, \( a \), \( b \), and \( c \) are constants taken from the terms of the quadratic equation. The solutions to the equation are the values of \( x \) that make the equation equal to zero.Using the quadratic formula involves plugging in our values of \( a \), \( b \), and \( c \), then carrying out the operations step-by-step to solve for \( x \). This formula guarantees the discovery of all potential solutions for a quadratic equation.

Discriminant

The discriminant is a key part of the quadratic formula, found within the square root, represented as \( b^2 - 4ac \). It provides valuable information about the nature of the solutions of a quadratic equation:

• If the discriminant is greater than zero, the quadratic equation has two distinct real solutions.
• If the discriminant is equal to zero, the equation has exactly one real solution, also known as a repeated root.
• If the discriminant is less than zero, the equation has no real solutions—only complex solutions.

In our problem, the equation \( x^2 + 5x - 6 = 0 \) has a discriminant of \( 49 \), which is greater than zero. Therefore, this means the equation has two distinct real solutions. This checking helps us know what kind of solutions to expect even before we start solving.

Real Solutions

Real solutions are the actual values that satisfy the quadratic equation. After determining the nature of the solutions using the discriminant, you can find the precise values by employing the quadratic formula. In our example, for the equation \( x^2 + 5x - 6 = 0 \), the discriminant is 49. Therefore, substituting \( a = 1 \), \( b = 5 \), and \( c = -6 \) into the quadratic formula, results in:\[x = \frac{-5 \pm 7}{2 \cdot 1}\]This calculation gives us two solutions:

• \( x_1 = 1 \)
• \( x_2 = -6 \)

These solutions are the points where the graph of the equation intersects the x-axis. Such solutions are real numbers, clear and definable within our number system, enhancing their applicability to real-world situations.

Verification of Solutions

Verification is the final step where we ensure the correctness of our solutions. This involves substituting the solutions back into the original equation to check if they indeed satisfy the equation.For our example equation \( x^2 + 5x - 6 = 0 \):

• Substitute \( x = 1 \): \( (1)^2 + 5(1) - 6 = 0 \), which simplifies to \( 0 \). Confirming \( x = 1 \) is a solution.
• Substitute \( x = -6 \): \( (-6)^2 + 5(-6) - 6 = 0 \), which also simplifies to \( 0 \). Confirming \( x = -6 \) is a solution.

Each solution must satisfy the original equation to be considered correct. By doing this check, we affirm that our solutions are valid and accurate, which reinforces the reliability of the quadratic formula method.