/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Find a quadratic equation with t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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. $$r_{1}=1+\sqrt{5}, r_{2}=1-\sqrt{5}$$

Short Answer

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

Step by step solution

01

Use the Form of a Quadratic Equation

A quadratic equation with roots \( r_1 \) and \( r_2 \) can be represented as \( (x - r_1)(x - r_2) = 0 \). In this case, the roots given are \( r_1 = 1 + \sqrt{5} \) and \( r_2 = 1 - \sqrt{5} \).
02

Multiply the Binomials

Using the roots from Step 1, set up the equation as \((x - (1 + \sqrt{5}))(x - (1 - \sqrt{5})) = 0\). Expand this expression: \( (x - 1 - \sqrt{5})(x - 1 + \sqrt{5}) = (x - 1)^2 - (\sqrt{5})^2 \).
03

Expand and Simplify

First, calculate \((x - 1)^2\): \((x - 1)^2 = x^2 - 2x + 1\). Then compute \( (\sqrt{5})^2 \): \(5\). Now plug these into the expanded product, giving \( x^2 - 2x + 1 - 5 \). Simplify this expression to get \( x^2 - 2x - 4 \).
04

Write the Final Quadratic Equation

The simplified result from Step 3 is our quadratic equation: \( x^2 - 2x - 4 = 0 \). This matches the required form with \( A = 1 \), \( B = -2 \), and \( C = -4 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Roots of a Quadratic Equation
Understanding the roots of a quadratic equation is fundamental when solving or forming these equations. The roots of a quadratic equation are the values of the variable that satisfy the equation, making the expression equal to zero. For a standard quadratic equation of the form \(Ax^2 + Bx + C = 0\), the roots \(r_1\) and \(r_2\) can often be found using the quadratic formula:
  • \(r_1, r_2 = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)
Alternately, if the roots are provided, like in the exercise where \(r_1 = 1 + \sqrt{5}\) and \(r_2 = 1 - \sqrt{5}\), the equation can be reconstructed. The equation can be expressed in factored form as \((x - r_1)(x - r_2) = 0\). This form directly uses the given roots to establish the equation's structure.
Knowing how to navigate between the roots and the expanded form of a quadratic equation is a critical skill for mathematical problem-solving in algebra.
Expanding Binomials
Expanding binomials is a key step to convert the factored form of a quadratic equation into its standard form. Binomials are simple expressions with two terms, like \((x - r)\). To form a full quadratic equation, you might need to expand the product of two binomials, especially useful when the roots of the quadratic are known.
For instance, consider the expression \((x - (1 + \sqrt{5}))(x - (1 - \sqrt{5}))\). Expanding these involves applying the formula for the product of two binomials:
  • \((x - 1 - \sqrt{5})(x - 1 + \sqrt{5}) = (x - 1)^2 - (\sqrt{5})^2\)
The rationale behind this formula is based on the difference of squares rule, \((a - b)(a + b) = a^2 - b^2\), simplifying the expansion process significantly. When properly simplified, this approach results in non-factored form, revealing the quadratic equation \(x^2 - 2x - 4 = 0\).
Mastering this technique is essential to simplifying complex equations and finding their roots effectively.
Standard Form of a Quadratic Equation
The standard form of a quadratic equation is vital in understanding and working with these mathematical structures. It is written as \(Ax^2 + Bx + C = 0\), where \(A\), \(B\), and \(C\) are constants with \(A eq 0\). This unified form is the foundation for both solving quadratic equations and analyzing their characteristics.
When deriving the quadratic equation from known roots, the process involves expanding binomials and simplifying the resultant expression. For the given example, the roots \(1 + \sqrt{5}\) and \(1 - \sqrt{5}\) were used to derive \(x^2 - 2x - 4 = 0\), which in the standard form is expressed with coefficients \(A = 1\), \(B = -2\), and \(C = -4\).
Writing equations in standard form allows for easy identification of the equation's key features, such as its roots, vertex, and axis of symmetry using established algebraic techniques. It's also the format that many mathematical operations and calculators are designed to handle most efficiently, making it universally important in algebraic computation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

First graph the two functions. Then use the method of successive approximations to locate, between successive thousandths, the \(x\)-coordinate of the point where the graphs intersect.Use a graphing utility to draw the graphs as well as to check your final answer. $$y=x^{3} ; y=4-x^{2}$$

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. $$\frac{5-x}{6 x^{2}-19 x+15}$$

Scipio Ferro of Bologna well-nigh thirty years ago discovered this rule and handed it on to Antonio Maria Fior of Venice, whose contest with Niccolò Tartaglia of Brescia gave Niccolò occasion to discover it. He / Tartaglial gave it to me in response to my entreaties, though withholding the demonstration. Armed with this assistance, I sought out its demonstration in /various / forms. - Girolamo Cardano, Ars Magna (Nuremberg, 1545 ) The quotation is from the translation of Ars Magna by T. Richard Witmer (New York: Dover Publications, 1993 ). In his book Ars Magna (The Great Art) the Renaissance mathematician Girolamo Cardano \((1501-1576)\) gave the following formula for a root of the equation \(x^{3}+a x=b\). $$x=\sqrt[3]{\frac{b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}-\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}$$ (a) Use this formula and your calculator to compute a root of the cubic equation \(x^{3}+3 x=76\) (b) Use a graph to check the answer in part (a). That is, graph the function \(y=x^{3}+3 x-76,\) and note the \(x-\) intercept. Also check the answer simply by substituting it in the equation \(x^{3}+3 x=76\)

(a) Let \(r_{1}, r_{2},\) and \(r_{3}\) be three distinct numbers that are roots of the equation \(f(x)=A x^{2}+B x+C=0 .\) Show that \(f(x)=0\) for all values of \(x .\) Hint: You need to show that \(A=B=C=0 .\) First show that both \(A\) and \(B\) are zero as follows. If either \(A\) or \(B\) were nonzero, then the equation \(f(x)=0\) would be a polynomial equation of degree at most 2 with three distinct roots. Why is that impossible? (b) Use the result in part (a) to prove the following identity: $$\begin{aligned}\frac{a^{2}-x^{2}}{(a-b)(a-c)} &+\frac{b^{2}-x^{2}}{(b-c)(b-a)} \\\&+\frac{c^{2}-x^{2}}{(c-a)(c-b)}-1=0 \end{aligned}$$ Hint: Let \(f(x)\) denote the quadratic expression on the left- hand side of the equation. Compute \(f(a), f(b)\) and \(f(c)\)

(a) Use a calculator to verify that the number \(\tan 9^{\circ}\) appears to be a root of the following equation: $$ x^{4}-4 x^{3}-14 x^{2}-4 x+1=0 $$ In parts (b) through (d) of this exercise, you will prove that \(\tan 9^{\circ}\) is indeed a root and that \(\tan 9^{\circ}\) is irrational. (b) Use the trigonometric identity $$ \tan 5 \theta=\frac{\tan ^{5} \theta-10 \tan ^{3} \theta+5 \tan \theta}{5 \tan ^{4} \theta-10 \tan ^{2} \theta+1} $$ to show that the number \(x=\tan 9^{\circ}\) is a root of the fifth-degree equation $$ x^{5}-5 x^{4}-10 x^{3}+10 x^{2}+5 x-1=0 $$Hint: In the given trigonometric identity, substitute \(\theta=9^{\circ}\) (c) List the possibilities for the rational roots of equation (2). Then use synthetic division and the remainder theorem to show that there is only one rational root. What is the reduced equation in this case? (d) Use your work in parts (b) and (c) to explain (in complete sentences) why the number \(\tan 9^{\circ}\) is an irrational root of equation (1).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.