Problem 18 - Elementary Algebra

Question

Solve the following quadratic equations. \((v+10)^{2}=121\)

Step-by-Step Solution

Answer

The solutions are \ v = 1 \ and \ v = -21 \.

Step 1 - Recognize the Equation Type

Identify that \( (v + 10)^{2} = 121 \) is a quadratic equation in the form \( (x + a)^2 = b \).

Step 2 - Take the Square Root of Both Sides

To eliminate the square, take the square root of both sides of the equation: \[ \sqrt{(v + 10)^{2}} = \sqrt{121}. \] This simplifies to \[ v + 10 = \pm 11. \]

Step 3 - Solve for Two Possible Values of v

The equation \( v + 10 = \pm 11 \) gives two possible equations: \[ v + 10 = 11 \] and \[ v + 10 = -11 \. \] Solving each one: \[ v + 10 = 11 \rightarrow v = 1 \] and \[ v + 10 = -11 \rightarrow v = -21. \]

Step 4 - Write the Final Solutions

The solutions to the equation \( (v + 10)^{2} = 121 \) are \ v = 1 \ and \ v = -21 \.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that involves a variable raised to the power of 2. In its standard form, a quadratic equation looks like this: \( ax^2 + bx + c = 0 \), where \(a\),\( b\), and \( c \) are constants. The highest exponent of the variable is 2, which makes it 'quadratic.' These equations are parabolas when graphed on a coordinate plane.

Quadratic equations can have different forms, including:

• Standard Form: \( ax^2 + bx + c = 0 \)
• Factored Form: \( a(x - r_1)(x - r_2) = 0 \)
• Vertex Form: \( a(x - h)^2 + k = 0 \)

The exercise you are solving is a quadratic equation in the special form \((x + a)^2 = b\), making it a bit easier to handle compared to the general quadratic form.

Square Roots

Square roots are a mathematical operation that reverses the process of squaring a number. In other words, the square root of \(x\) (written as \(\sqrt{x}\)) is a number \( y\) such that \( y^2 = x \). In the context of quadratic equations, taking the square root of both sides is a crucial step.

In the given problem, we have \((v + 10)^2 = 121\). To clear the square, we need to take the square root of both sides:

• \( \sqrt{(v + 10)^2} = \sqrt{121} \)

This simplifies to \( v + 10 = \pm 11 \), introducing two separate linear equations for us to solve. Remember, taking the square root of both sides introduces a positive and negative solution (represented by \(\pm\)).

Solving Equations

Solving equations involves finding the values for the variables that make the equation true. Let's focus on how to solve the simplified equations from your exercise.

After taking the square root, we had two equations:

• \(v + 10 = 11 \)
• \(v + 10 = -11 \)

To solve these, isolate the variable \(v\) by performing inverse operations:

For \(v + 10 = 11\):

• Subtract 10 from both sides.

\(v = 1\)

For \(v + 10 = -11\):

• Subtract 10 from both sides.

\(v = -21\)

Thus, the solutions to the quadratic equation are \(v = 1\) and \(v = -21\).

Equation Types

Equations can come in many types beyond just quadratic. Knowing what type of equation you're dealing with can guide you on the best approach to solve it.

Some equation types include:

• Linear Equations: \(ax + b = 0\)
• Quadratic Equations: \(ax^2 + bx + c = 0\)
• Exponential Equations: \(a^x = b\)
• Logarithmic Equations: \( \log_a(x) = b\)
• Trigonometric Equations: \( \sin(x) = c\), \( \cos(x) = c\)

Any quadratic equation can be solved using methods like:

• Factoring
• Using the Quadratic Formula: \(\frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a}\)
• Completing the Square

In the exercise, we used the 'taking the square root' method to solve the quadratic equation efficiently.