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Chapter 12: Problem 18

Find a quadratic equation with rational coefficients, one of whose roots is the given number. Write your answer so that the coefficient of \(x^{2}\) is 1. Use either of the methods shown in Example 3 $$r_{1}=2-\sqrt{3}$$

Short Answer

Expert verified
The quadratic equation is \( x^2 - 4x + 1 = 0 \).

Step by step solution

01

Recognize Root Pairs

Since the coefficients of the quadratic are rational, complex roots occur in conjugate pairs. Given one root \( r_1 = 2 - \sqrt{3} \), the other root \( r_2 \) must be its conjugate: \( r_2 = 2 + \sqrt{3} \).
02

Use Sum and Product of Roots

For roots \( r_1 \) and \( r_2 \), a quadratic equation can be expressed as \( x^2 - (r_1 + r_2)x + r_1r_2 = 0 \).
03

Calculate the Sum of the Roots

The sum of the roots \( r_1 + r_2 \) is \((2 - \sqrt{3}) + (2 + \sqrt{3}) = 4\).
04

Calculate the Product of the Roots

The product of the roots \( r_1 \times r_2 \) is \((2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1\).
05

Formulate the Quadratic Equation

Plugging the sum and product into the equation form, we have \( x^2 - 4x + 1 = 0 \). This is the quadratic equation with the rational coefficients where one root is \( 2 - \sqrt{3} \).

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Key Concepts

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

Rational Coefficients
When dealing with quadratic equations, rational coefficients mean that the numbers in front of each term must be either whole numbers or fractions, essentially any number that can be expressed as a ratio of two integers.
This is important as it limits the type of roots the equation can have.
  • If a root of the quadratic equation is irrational, such as involving a square root, the quadratic must have its conjugate as the other root.
  • This ensures that when forming the equation, any irrational parts cancel out, leaving only rational terms.
For example, if you have a root like \(2 - \sqrt{3}\), its conjugate \(2 + \sqrt{3}\) must also be a root to maintain rationality in the coefficients.
Roots of Polynomials
The root of a polynomial refers to any value that, when substituted for the variable, makes the polynomial equal to zero.
So, for a quadratic equation of the form \(ax^2 + bx + c = 0\), the solutions for \(x\) are the roots.Quadratics, being polynomials of degree 2, have exactly two roots. These roots can be found using several methods such as factoring, completing the square, or the quadratic formula.
However, when specific roots are given, like in our problem, an understanding of their properties, such as conjugates, is often required.
  • The case of having one root \(2 - \sqrt{3}\) prompts the use of its conjugate \(2 + \sqrt{3}\) as the other root because their product and sum will render rational results necessary for coefficients of the equation.
Sum and Product of Roots
Realizing how sums and products of the roots relate to a quadratic equation is a handy tool.
For any quadratic equation in the form \(x^2 + bx + c = 0\), where 1 is the leading coefficient:
  • The sum of the roots \(r_1 + r_2\) is equal to \(-b\).
  • The product of the roots \(r_1 \cdot r_2\) is equal to \(c\).
So, when we deduce a quadratic based on its roots, we use these relationships to form it.
For the roots \(2 - \sqrt{3}\) and \(2 + \sqrt{3}\), the sum and product calculations were as follows: the sum was 4 and the product was 1.
These were used in the standard form equation as \(x^2 - 4x + 1 = 0\).
This process demonstrates how the interplay of sum and product of roots with the quadratic coefficients ensures that the equation remains reliable and rational.

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Most popular questions from this chapter

Determine a quadratic equation with the given roots: (a) \(a / b,-b / a\) (b) \(-a+2 \sqrt{2 b},-a-2 \sqrt{2 b}\)

\(Determine the partial fraction decomposition for each of the given rational expressions. Hint: In Exercises \)17,18,\( and \)26,\( use the rational roots theorem to factor the denominator. \)\frac{2 x+1}{x^{3}-5 x}$$

Find a quadratic equation with the given roots. Write your answers in the form \(A x^{2}+B x+C=0\) Suggestion: Make use of Table 2. $$r_{1}=1+\sqrt{5}, r_{2}=1-\sqrt{5}$$

This exercise outlines a proof of the rational roots theorem. At one point in the proof, we will need to rely on the following fact, which is proved in courses on number theory. FACT FROM NUMBER THEORY Suppose that \(A, B\), and \(C\) are integers and that \(A\) is a factor of the number \(B C .\) If \(A\) has no factor in common with \(C\) (other than 卤1 ), then \(A\) must be a factor of \(B\) (a) Let \(A=2, B=8,\) and \(C=5 .\) Verify that the fact from number theory is correct here. (b) Let \(A=20, B=8,\) and \(C=5 .\) Note that \(A\) is a factor of \(B C,\) but \(A\) is not a factor of \(B .\) Why doesn't this contradict the fact from number theory? (c) Now we're ready to prove the rational roots theorem. We begin with a polynomial equation with integer coefficients: \(-a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}=0 \quad\left(n \geq 1, a_{n} \neq 0\right)\) We assume that the rational number \(p / q\) is a root of the equation and that \(p\) and \(q\) have no common factors other than \(1 .\) Why is the following equation now true? \(a_{n}\left(\frac{p}{q}\right)^{n}+a_{n-1}\left(\frac{p}{q}\right)^{n-1}+\cdots+a_{1}\left(\frac{p}{q}\right) a_{0}=0\) (d) Show that the last equation in part (c) can be written \(p\left(a_{n} p^{n-1}+a_{n-1} q p^{n-2}+\cdots+a_{1} q^{n-1}\right)=-a_{0} q^{n}\) since \(p\) is a factor of the left-hand side of this last equation, \(p\) must also be a factor of the right-hand side. That is, \(p\) must be a factor of \(a_{0} q^{n}\). But since \(p\) and \(q\) have no common factors, neither do \(p\) and \(q^{n}\) Our fact from number theory now tells us that \(p\) must be a factor of \(a_{0}\), as we wished to show. (The proof that \(q\) is a factor of \(a_{n}\) is carried out in a similar manner.)

This exercise completes the discussion of improper rational expressions in this section. (a) Use long division to obtain the following result: \(\frac{2 x^{3}+4 x^{2}-15 x-36}{x^{2}-9}=(2 x+4)+\frac{3 x}{x^{2}-9}\) (b) Find constants \(A\) and \(B\) such that \(3 x /\left(x^{2}-9\right)=A /(x-3)+B /(x+3) .\) (According to the text, you should obtain \(A=B=3 / 2 .\) )
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