Problem 16 - Algebra 2

Question

Write a quadratic equation in standard form with the given roots. \(-6,-8\)

Step-by-Step Solution

Answer

The quadratic equation is \( x^2 + 14x + 48 = 0 \).

Step 1: Understand the problem

We need to write a quadratic equation in the standard form that has the given roots. The roots provided are -6 and -8.

Step 2: Use the root form of a quadratic equation

Recall that if a quadratic equation has roots \( r_1 \) and \( r_2 \), it can be expressed as \( (x - r_1)(x - r_2) = 0 \). For the given roots -6 and -8, the equation becomes \( (x + 6)(x + 8) = 0 \).

Step 3: Expand the equation

Now, expand \( (x + 6)(x + 8) \) using the distributive property: \[(x + 6)(x + 8) = x^2 + 8x + 6x + 48.\] Combine like terms to simplify.

Step 4: Simplify the expression

Combine the like terms from the expansion: \( x^2 + 8x + 6x + 48 = x^2 + 14x + 48 \).

Step 5: Write the equation in standard form

The simplified expression \( x^2 + 14x + 48 \) is the quadratic equation in standard form. Thus, the equation is \( x^2 + 14x + 48 = 0 \).

Key Concepts

Roots of Quadratic Equations

The roots of a quadratic equation are the numbers that satisfy the equation when it is set to zero. These roots are often referred to as solutions or zeros. In a quadratic equation of the form \[ ax^2 + bx + c = 0, \]the roots are the values of \( x \) that make the equation true. When given specific roots, like -6 and -8 in our example, you can easily form a quadratic equation by reversing the process of finding roots.To form an equation from specific roots \( r_1 \) and \( r_2 \), use the root form, which states:\[ (x - r_1)(x - r_2) = 0. \]For -6 and -8, plug these values in

• \( r_1 = -6 \) becomes \( (x + 6) \)
• \( r_2 = -8 \) becomes \( (x + 8) \)

Multiplying these expressions gives a quadratic equation in factored form, before it's expanded and put in standard form.

Standard Form of Quadratic Equations

The standard form of a quadratic equation is expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. This form is useful because it allows the equation to be easily analyzed and solved using various methods, such as factoring, completing the square, or using the quadratic formula.In our example, once the factored form \((x + 6)(x + 8) = 0\) is expanded through multiplication, the result is:\[ x^2 + 14x + 48 = 0. \]Here,

• \( a = 1 \) (the coefficient of \( x^2 \))
• \( b = 14 \) (the coefficient of \( x \))
• \( c = 48 \) (the constant term)

The standard form makes it easy to identify the coefficients and understand the behavior of the quadratic equation under various conditions.

Distributive Property

The distributive property is a fundamental algebraic principle used to expand expressions. It allows you to multiply a single term by two or more terms inside a set of parentheses. In general, the distributive property is expressed as:\[ a(b + c) = ab + ac. \]This property is essential when expanding quadratic equations, which are initially written in factored form.For the given problem, using the distributive property helps convert \((x + 6)(x + 8)\) into the standard form. Here's how it applies:

• First, multiply \( x \) by both \( x \) and \( 8 \): \( x(x + 8) = x^2 + 8x \).
• Then, multiply \( 6 \) by both \( x \) and \( 8 \): \( 6(x + 8) = 6x + 48 \).
• Combine these results to get the expanded form: \( x^2 + 8x + 6x + 48 \).

Finally, by combining like terms, you arrive at the standard form \( x^2 + 14x + 48 = 0 \). The distributive property is a powerful tool for working with equations of all kinds, but especially quadratic equations.