Problem 97 - Precalculus Student Solutions Manual 5th

Question

Find the values of \(a, b,\) and \(c\) for which the quadratic equation $$ a x^{2}+b x+c=0 $$ has the given numbers as solutions. (Hint: Use the zero-factor property in reverse.) $$1+\sqrt{2}, 1-\sqrt{2}$$

Step-by-Step Solution

Answer

The values are \(a = 1\), \(b = -2\), and \(c = -1\).

Step 1: Understand the given solutions

The given solutions of the quadratic equation are \(1+\sqrt{2}\) and \(1-\sqrt{2}\).

Step 2: Use the zero-factor property

According to the zero-factor property, if the roots of the quadratic equation \(ax^2 + bx + c = 0\) are \(r_1\) and \(r_2\), then the equation can be written as \(a(x - r_1)(x - r_2) = 0\). In this problem, \(r_1 = 1 + \sqrt{2}\) and \(r_2 = 1 - \sqrt{2}\).

Step 3: Set up the quadratic equation

Substitute the given roots into the equation form: \(a(x - (1+\sqrt{2}))(x - (1-\sqrt{2})) = 0\).

Step 4: Simplify the equation

Expand the product: \((x - (1+\sqrt{2}))(x - (1-\sqrt{2})) = (x - 1 - \sqrt{2})(x - 1 + \sqrt{2})\). Use the difference of squares formula: \((x - 1)^2 - (\sqrt{2})^2\).

Step 5: Further simplification

Perform the operations: \((x - 1)^2 - 2 = x^2 - 2x + 1 - 2 = x^2 - 2x - 1\).

Step 6: Identify coefficients

From the expanded equation \(a(x^2 - 2x - 1) = 0\), it can be seen that the equation matches the form \(a x^2 + b x + c = 0\) with coefficients identified as: \(a = 1\), \(b = -2\), and \(c = -1\).

Key Concepts

zero-factor property

The zero-factor property is a key concept when dealing with quadratic equations. It states that if a product of two factors is zero, then at least one of the factors must be zero. This property is used to solve quadratic equations by factoring.
For example, consider a quadratic equation with roots (solutions) \(r_1\) and \(r_2\). According to the zero-factor property:

• The equation can be written as \a(x - r_1)(x - r_2) = 0\.
• Here, \(r_1\) and \(r_2\) are the solutions to the equation \ax^2 + bx + c = 0\.

It's crucial to understand this property because it helps break down complex problems into simpler parts that are easier to manage. By using this property in reverse, as hinted in the problem, we can form a quadratic equation from its given roots and proceed to find its coefficients.

difference of squares

The difference of squares is another fundamental algebraic concept used to simplify expressions. This formula states that:
\ (a^2 - b^2 = (a - b)(a + b)) \
It helps factorize expressions where one term is squared minus another squared term.
In our problem, we use the difference of squares to simplify the expression \( (x - 1 - \sqrt{2})(x - 1 + \sqrt{2}) \):

• Rewrite the expression as a difference of squares: \ (x - 1)^2 - (\sqrt{2})^2 \.
• Perform the operations to get \ x^2 - 2x + 1 - 2 = x^2 - 2x - 1 \.

The difference of squares helps us simplify complex-looking expressions and make our calculations more manageable.

solving quadratic equations

Solving quadratic equations is a vital skill in algebra. Quadratic equations are in the form \a x^2 + b x + c = 0\. Here are the basic steps to solve them:

• Identify the roots (solutions) of the equation. These are values of \x\ that make the equation true.
• Use the zero-factor property to express the equation in factored form: \ a(x - r_1)(x - r_2) = 0 \.
• Simplify the equation, often using techniques like the difference of squares.
• Expand and combine like terms to identify the coefficients \a\, \b\, and \c\.

In our example, the given solutions are \(1+\sqrt{2}\) and \(1-\sqrt{2}\). Applying the zero-factor property and the difference of squares, we simplify the expression to \x^2 - 2x - 1\. Thus, the coefficients are \(a = 1, b = -2, c = -1\). This step-by-step approach helps us grasp how to solve quadratic equations systematically.