Problem 28 - Elementary Algebra

Question

\(v^{2}-16 v+15=0\)

Step-by-Step Solution

Answer

The solutions are \( v = 15 \) and \( v = 1 \).

Step 1 - Identify the equation form

The given equation is a quadratic equation in standard form: \( v^2 - 16v + 15 = 0 \)

Step 2 - Apply the quadratic formula

The quadratic formula is \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Here, \( a = 1 \), \( b = -16 \), and \( c = 15 \).

Step 3 - Compute the discriminant

Calculate the discriminant \( b^2 - 4ac \): \( (-16)^2 - 4(1)(15) = 256 - 60 = 196 \).

Step 4 - Find the square root of the discriminant

The square root of the discriminant is \( \sqrt{196} = 14 \).

Step 5 - Solve for \( v \)

Substitute back into the quadratic formula: \( v = \frac{-(-16) \pm 14}{2(1)} \).This simplifies to two solutions: \( v = \frac{16 + 14}{2} = 15 \) and \( v = \frac{16 - 14}{2} = 1 \).

Key Concepts

quadratic formula

Solving quadratic equations is simple once you know the quadratic formula. The quadratic formula is:
\( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
It might look complicated, but it’s really just plug-and-play. You just need to know your values of a, b, and c from the quadratic equation format:
\( ax^2 + bx + c = 0 \)
In this example, we have:

• a = 1
• b = -16
• c = 15

When you substitute these values into the formula, you then solve using basic arithmetic steps. Remember, the quadratic formula helps you find the values of the variable (in this case, v) that make the equation true.

discriminant

The discriminant is a very important part of the quadratic formula. It tells us if the roots of our quadratic equation are real or complex numbers.
The discriminant is found inside the square root of the quadratic formula:
\( \sqrt{b^2 - 4ac} \)
In our exercise:

• b = -16
• a = 1
• c = 15

So, we calculate:
\( (-16)^2 - 4(1)(15) = 256 - 60 = 196 \)
A positive discriminant (like in this case, 196) means we have two real and distinct roots. A zero discriminant means the equation has one real root (a repeated root), and a negative discriminant indicates two complex roots.

roots of quadratic equations

Finding the roots of a quadratic equation is the goal of using the quadratic formula. Once we have calculated the discriminant, we use it in the quadratic formula to find our roots.
For the example problem, the discriminant was 196. We need to find its square root:
\( \sqrt{196} = 14 \)
Now we substitute it back into the formula:
\( v = \frac{-(-16) \pm 14}{2(1)} \)
This simplifies to two solutions:

• First root: \( v = \frac{16 + 14}{2} = 15 \)
• Second root: \( v = \frac{16 - 14}{2} = 1 \)

These values of v are called the roots of the quadratic equation. They are the values that make the original equation true.

algebra steps

Understanding the algebra steps involved in solving quadratic equations is crucial. Here’s a brief recap of the steps followed in our exercise:
1. **Identify the quadratic equation** — Make sure the equation is in the standard format \( ax^2 + bx + c = 0 \).
2. **Write down the quadratic formula** — \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
3. **Find a, b, and c** — From the equation \( v^2 -16v + 15 = 0 \), we have:

• a = 1
• b = -16
• c = 15

4. **Calculate the discriminant** — \( b^2 - 4ac \). Substitute the values and evaluate.
5. **Find the square root of the discriminant** — Simplify \( \sqrt{196} = 14 \).
6. **Solve for the roots using the quadratic formula**.
Each step depends on accurate arithmetic and careful substitution. Practice these steps, and quadratic equations will become much easier to solve!