Problem 22 - Beginning and Intermediate Algebra

Question

Solve each equation. $$z^{2}-12 z+11=0$$

Step-by-Step Solution

Answer

The solutions to the equation \(z^2 - 12z + 11 = 0\) are \(z = 11\) and \(z = 1\).

Step 1: Attempt Factoring

Let's attempt to factor the given quadratic equation, \(z^2 - 12z + 11 = 0\). We are looking for two numbers that multiply to 11 and add up to -12. Upon inspection, we can see that the two numbers we're looking for are -11 and -1. So, we can factor the quadratic as follows: \((z - 11)(z - 1) = 0\)

Step 2: Solve for z

Now that we have factored the equation, we can set each factor equal to 0 and solve for \(z\). 1. \(z - 11 = 0\) Adding 11 to both sides: \(z = 11\) 2. \(z - 1 = 0\) Adding 1 to both sides: \(z = 1\)

Step 3: State the Solution

The solutions to the given quadratic equation \(z^2 - 12z + 11 = 0\) are \(z = 11\) and \(z = 1\).

Key Concepts

Factoring Quadratics

Quadratic equations, like the one given in the exercise, generally take the form of \( ax^2 + bx + c = 0 \). A key technique for solving these is factoring. Factoring involves breaking down the quadratic equation into simpler expressions called factors, which when multiplied together give the original equation. In our specific example, \( z^2 - 12z + 11 = 0 \), the goal is to express it as \((z - p)(z - q) = 0\). Here, \( p \) and \( q \) are numbers such that their product equals 11 (the constant term) and their sum equals -12 (the coefficient of the middle term \( z \)).

• Find two numbers that multiply to give the constant term: 11.
• These numbers also need to add up to the coefficient of the linear term: -12.
• In this case, the numbers are -11 and -1.

Thus, the equation \( z^2 - 12z + 11 \) can be rewritten in factored form as \((z - 11)(z - 1) = 0\). This factoring is crucial since it allows us to find the equation's roots in the next steps.

Roots of Equations

Once the quadratic equation is factored, we use the zero product property to find its roots or solutions.The zero product property states that if the product of two factors is zero, then at least one of the factors must be zero. This means that solving \((z - 11)(z - 1) = 0\) requires each factor to be set to zero, leading to two mini-equations:

• \( z - 11 = 0 \)
• \( z - 1 = 0 \)

We solve each equation individually:

• For \( z - 11 = 0 \), add 11 to both sides: \( z = 11 \).
• For \( z - 1 = 0 \), add 1 to both sides: \( z = 1 \).

Therefore, the roots of the equation, which are the values of \( z \) that satisfy the original equation, are \( z = 11 \) and \( z = 1 \). Finding these roots provides the solutions to the quadratic equation.

Polynomial Equations

Quadratic equations are a subset of polynomial equations, which encompass any expression involving powers of a variable. Let's delve a bit into this broader category.A polynomial equation can have terms involving powers of a variable, where each term contains a coefficient, a variable, and an exponent. These can have varying exponents and any amount of terms, with the general form \( a_n x^n + a_{n-1} x^{n-1} + \,...\, + a_1 x + a_0 = 0 \) where \( a_n,...,a_0 \) are constants.

• A linear equation, like \( ax + b = 0 \), is a polynomial of degree 1.
• A quadratic equation, \( ax^2 + bx + c = 0 \), is a polynomial of degree 2.
• Polynomial equations can be more complex, having degrees 3 (cubic), 4 (quartic), etc.

Solving polynomial equations often involves factoring, especially for lower-degree polynomials like quadratics. Higher-degree polynomials might require different techniques, such as synthetic division or using the Rational Root Theorem.