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Chapter 5: Problem 8

The number 9 is a double solution of the following equation. Why is this so? $$ (x-9)^{2}=0 $$

Short Answer

Expert verified
The number 9 is a double solution because it satisfies the equation (x-9)^{2}=0 and has a multiplicity of 2.

Step by step solution

01

Understand the Equation

Begin by examining the given equation (x-9)^{2}=0.
02

Identify the Root

Set the inside of the squared term equal to zero to find the root: x-9=0 then solve for x. This results in x=9.
03

Determine the Multiplicity

Since the root appears inside a squared term, it has a multiplicity of 2. This means that the root 9 is repeated twice. Mathematically, (x-9)(x-9) = 0 implies that 9 is a solution that appears twice.
04

Conclude the Double Solution

Therefore, the number 9 is a double solution of the equation (x-9)^{2}=0 because it satisfies the equation and appears with a multiplicity of 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multiplicity
Multiplicity refers to the number of times a particular solution, or root, satisfies an equation.
In the context of quadratic equations, if a root has a multiplicity greater than 1, it means the root is repeated.
For example, in the equation \((x-9)^2=0\), the root 9 has a multiplicity of 2.
This is because the factor \(x-9\) is squared, implying that 9 is a solution that is counted twice.
Understanding multiplicity helps in identifying how roots behave and how they are represented graphically.
A root with high multiplicity means the graph touches the x-axis at this point but does not cross it.
Solving Quadratic Equations
Quadratic equations are polynomials of degree 2 and generally take the form of \(ax^2 + bx + c = 0\).
There are several methods to solve quadratic equations:
  • Factoring: Express the quadratic as a product of binomials and solve for the roots.
  • Completing the Square: Rewrite the quadratic in the form \((x-h)^2 = k\) and then solve for x.
  • Quadratic Formula: Use the formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) to find the roots directly.
In the given exercise, the equation is already factored as \((x-9)^2=0\).
To solve it, we set the expression inside the parentheses equal to zero:
\(x-9=0\) and then solve for \(x\) to get \(x=9\).
This provides us with the root of the equation.
Roots of Equations
Roots of an equation are the values of a variable that satisfy the equation, making it true.
For a quadratic equation of the form \(ax^2 + bx + c = 0\), these roots are the solutions for \(x\).
Roots can be real or complex, depending on the discriminant \(b^2 - 4ac\).
In the given equation \((x-9)^2 = 0\), solving gives us the root \(x = 9\).
The unique aspect here is that 9 is a double root, meaning it has a multiplicity of 2.
This multiplicity implies that the parabola touches the x-axis at \(x = 9\) but does not pass through it.
Recognizing double roots is crucial for fully understanding an equation's solutions and visualizing their geometric representation.

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Most popular questions from this chapter

Factor each polynomial completely. \(25 x^{2}-10 x y+y^{2}-36 z^{2}\)

Solve each problem. Find three consecutive odd integers such that 3 times the sum of all three is 18 more than the product of the first and second integers.

Identify each monomial as a perfect square, a perfect cube, both of these, or neither of these. (a) \(4 x^{3}\) (b) \(8 y^{6}\) (c) \(49 x^{12}\) (d) \(81 r^{10}\) (e) \(64 x^{6} y^{12}\) (f) \(125 t^{6}\)

Factor each trinomial. \(-2 a^{2}-5 a b-2 b^{2}\)

Factor each binomial completely. \(27 w^{3}-216 z^{3}\)
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