Problem 75 - Elementary Algebra

Question

Solve. $$ x_{2}+5 x-50=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 5\) and \(x = -10\).

Step 1: Convert to Standard Quadratic Form

The equation given is already in the standard quadratic form: \(x^2 + 5x - 50 = 0\). This form is \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = 5\), and \(c = -50\).

Step 2: Identify Coefficients

For the quadratic equation \(x^2 + 5x - 50 = 0\), identify the coefficients: \(a = 1\), \(b = 5\), and \(c = -50\).

Step 3: Apply Quadratic Formula

Use the quadratic formula to solve for \(x\): \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute the coefficients \(a\), \(b\), and \(c\) into the formula: \[x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-50)}}{2 \cdot 1}\]

Step 4: Simplify the Discriminant

Calculate the discriminant: \[b^2 - 4ac = 5^2 - 4 \times 1 \times (-50) = 25 + 200 = 225\]

Step 5: Calculate the Roots

Now that we know the discriminant \( b^2 - 4ac = 225 \), plug it back into the formula:\[x = \frac{-5 \pm \sqrt{225}}{2}\]Since \(\sqrt{225} = 15\), the expression becomes: \[x = \frac{-5 \pm 15}{2}\]

Step 6: Find Possible Values of x

Solve for the two possibilities:1. \(x = \frac{-5 + 15}{2} = \frac{10}{2} = 5\)2. \(x = \frac{-5 - 15}{2} = \frac{-20}{2} = -10\).

Key Concepts

Quadratic Formula

To solve any quadratic equation, we can use the powerful quadratic formula. The quadratic formula is given by:

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

This formula helps find the solutions or "roots" of a quadratic equation, if they exist.
We replace \(a\), \(b\), and \(c\) in the formula with their values from the given equation to find the value of \(x\). Each quadratic equation can have two solutions, which is why we see "plus or minus" in the formula.
This means we need to calculate for both \(+\) and \(-\) to get the two potential roots.

Standard Quadratic Form

The standard quadratic form is a way of writing quadratic equations so that they are easily identifiable and solvable.
It is expressed as:

• \( ax^2 + bx + c = 0 \)

Here, \(a\), \(b\), and \(c\) are constants, with \(a\) not equal to zero. This format is crucial because it allows us to apply various methods, such as factoring or using the quadratic formula, to solve the equation.
In our example equation, \(x^2 + 5x - 50 = 0\), it is already in standard quadratic form with \(a = 1\), \(b = 5\), and \(c = -50\). Knowing these values is the first step in solving the equation using the quadratic formula.

Discriminant

The discriminant is a key component of the quadratic formula, found inside the square root. Its formula is:

• \( b^2 - 4ac \)

Calculating the discriminant helps us understand the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root, also known as a repeated or double root.
• If it is negative, there are no real roots, but two complex roots.

In our case, the discriminant is 225:

• \( 5^2 - 4 \times 1 \times (-50) = 225 \)

Since 225 is positive, we know there are two distinct real solutions for our equation.

Solving Quadratic Equations

Solving quadratic equations involves finding the value(s) of \(x\) that satisfy the equation.
There are several methods to solve these equations, such as factoring, completing the square, or using the quadratic formula. Many times, the quadratic formula is the most convenient method, especially when the equation cannot be factored easily or when it involves complex numbers.
Let’s revisit our problem:

• Initial equation: \(x^2 + 5x - 50 = 0\)
• Quadratic formula applied: \( x = \frac{-5 \pm 15}{2} \)
• Two solutions found: \( x = 5 \) and \( x = -10 \)

By finding these roots, we can say that these values of \(x\) satisfy the original quadratic equation, making them the solutions.