Problem 81 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find two quadratic equations having the given solutions. (There are many correct answers.) $$-6,5$$

Step-by-Step Solution

Answer

The two quadratic equations which have roots of -6 and 5 are: \(x^{2}+x-30=0\) and \(2x^{2}+2x-60=0\).

Step 1: Determine the Sums and Products of the Roots

From the problem, we know that the roots are -6 and 5. So we first determine their sum and product: The sum of the roots (-6 and 5) is -6+5=-1, and the product of the roots is -6*5=-30.

Step 2: Create the First Quadratic Equation

We'll now form the quadratic equation with the sum and product of roots. The general form of a quadratic equation is \(x^{2} - (sum \: of\: the\: roots )\cdot x + (product\: of \:the\: roots) = 0\). Replacing the sum of roots by -1 and product of roots by -30, we get \(x^{2} - (-1)\cdot x - 30=0\), which simplifies to \(x^{2}+x-30=0\).

Step 3: Create the Second Quadratic Equation

Changing the coefficients of the equation will result in another valid quadratic equation with the same roots. Therefore, we can multiply all coefficients of the train quadratic equation by a non-zero constant, for example 2. Doing this, we will get a second quadratic equation: \(2x^{2}+2x-60=0\).

Key Concepts

Sum of Roots

The sum of the roots of a quadratic equation is an important concept when working with polynomial expressions. Specifically, for a quadratic equation of the form \(ax^2 + bx + c = 0\), the sum of its roots (\(\alpha + \beta\)) is

• represented by the formula \(-\frac{b}{a}\).

This means when we know the roots of the equation, calculating their sum helps us understand more about the structure of the equation.
In our example, with roots \(-6\) and \(5\), the sum is calculated as:

• \(-6 + 5 = -1\).

This sum helps in forming the quadratic equation by determining the coefficient of the linear term (\(-bx\)) in the equation.

Product of Roots

The product of the roots of a quadratic equation is equally significant. It adds another layer of understanding to the relationships within the quadratic equation. For a standard quadratic formula \(ax^2 + bx + c = 0\), the product of its roots (\(\alpha \beta\)) can be found using the formula:

• \(\frac{c}{a}\).

Knowing the product is particularly useful for checking your work or forming equations. In our exercise, the root values \(-6\) and \(5\) lead us to a product calculated as:

• \(-6 \times 5 = -30\).

This product determines the constant term (\(c\)) in the quadratic equation. Understanding both sum and product allows us to recreate the original quadratic equation.

Forming Quadratic Equations

Creating a quadratic equation from known roots is an essential skill in algebra. Once you've figured out the sum and product of the roots, you can form different yet mathematically equivalent quadratic equations.
We start with the known general form:

• \(x^2 - (\text{sum of the roots})x + (\text{product of the roots}) = 0\).

Using our example with roots \(-6\) and \(5\), we replace the sum and product in the equation:

• \(x^2 - (-1)x - 30 = 0\), which simplifies to \(x^2 + x - 30 = 0\).

Quadratic equations can have many forms by multiplying through by a constant. For instance, by multiplying the entire equation \(x^2+x-30=0\) by \(2\), we get:

• \(2x^2 + 2x - 60 = 0\).

Each form remains valid and has the same roots, highlighting flexibility within quadratic expressions.