/*! 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 147 If roots of the equation \(x^{2}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 147

If roots of the equation \(x^{2}-b x+c=0\) be two consectutive integers, then \(b^{2}-4 c\) equals \(\quad\) [2005] (A) \(-2\) (B) 3 (C) 2 (D) 1

Short Answer

Expert verified
The correct answer is (D) 1.

Step by step solution

01

Recognize the Nature of Roots

The problem states that the two roots are consecutive integers. Let's denote these roots as \(n\) and \(n+1\).
02

Write the Sum and Product of Roots

According to Vieta's formulas, for a quadratic equation \(x^2 - bx + c = 0\), the sum of the roots \(n + (n+1)\) is equal to \(b\). Therefore, \(b = 2n + 1\). Similarly, the product of the roots \(n(n+1)\) is equal to \(c\). Thus, \(c = n(n+1)\).
03

Calculate the Difference \(b^2 - 4c\)

Substitute \(b = 2n + 1\) and \(c = n(n+1)\) into the expression \(b^2 - 4c\):\[b^2 - 4c = (2n + 1)^2 - 4(n(n+1))\]Expand and simplify:\[(2n + 1)^2 = 4n^2 + 4n + 1\]\[4(n(n+1)) = 4n^2 + 4n\]The expression becomes:\[4n^2 + 4n + 1 - 4n^2 - 4n = 1\]
04

Conclude the Solution

Since all terms cancel out except for \(1\), we have calculated that \(b^2 - 4c = 1\).

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 91影视!

Key Concepts

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

Quadratic Equation
A quadratic equation is a second-order polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants and \( a eq 0 \). In this context, the quadratic equation is given as \( x^2 - bx + c = 0 \). Quadratic equations often have two solutions or roots, which can either be real or complex numbers.
To find the roots of a quadratic equation, we can use the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
The term \( b^2 - 4ac \) is called the discriminant and it determines the nature of the roots. In our exercise, it's important because we're specifically working with the concept of roots being consecutive integers.
Vieta's Formulas
Vieta's formulas are a valuable tool when dealing with quadratic equations. They relate the coefficients of the equation to sums and products of its roots, without actually needing to solve the equation.
For a quadratic equation of the form \( x^2 - bx + c = 0 \), Vieta's formulas provide:
  • The sum of the roots \( (r_1 + r_2) = b \)
  • The product of the roots \( r_1 \cdot r_2 = c \)
In our exercise, we apply Vieta's formulas by considering the roots as consecutive integers \( n \) and \( n+1 \). Therefore, their sum is \( n + (n+1) = 2n + 1 \), which equals \( b \), and their product \( n(n+1) = n^2 + n \), which equals \( c \). This insight allows us to express \( b \) and \( c \) in terms of \( n \), leading to the simplification of the expression \( b^2 - 4c \).
Roots and Coefficients Relationship
Understanding the relationship between the roots of a polynomial and its coefficients is crucial for solving and simplifying such problems.
In a quadratic equation such as \( x^2 - bx + c = 0 \), the coefficients \( b \) and \( c \) are directly related to its roots via Vieta's formulas. This means:
  • The sum of the roots is equal to the coefficient \( b \), allowing us to identify patterns like consecutive integers forming the roots.
  • The product of the roots gives us the constant term \( c \).
In this exercise, recognizing that the roots are consecutive integers helps simplify the existing equation. This particular relationship enables us to substitute \( n \) into expressions for \( b \) and \( c \), simplifying \( b^2 - 4c \) down to 1. It beautifully demonstrates the power of using algebraic structure and relationships to solve problems efficiently.

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

If \(\alpha, \beta\) are the roots of the equation \(a x^{2}+b x+c=0\), \((a \neq 0)\) and \(\alpha+\delta, \beta+\delta\) are the roots of \(A x^{2}+B x+\) \(C=0,(A \neq 0)\) for some constant \(\delta\), then (A) \(\frac{b^{2}-4 a c}{a^{2}}=\frac{B^{2}-4 A C}{A^{2}}\) (B) \(\frac{b^{2}-2 a c}{a^{2}}=\frac{B^{2}-2 A C}{A^{2}}\) (C) \(\frac{b^{2}-8 a c}{a^{2}}=\frac{B^{2}-8 A C}{A^{2}}\) (D) None of these

The solution set of \((x)^{2}+(x+1)^{2}=25\), where \((x)\) is the least integer greater than or equal to \(x\), is (A) \((2,4)\) (B) \((-5,4] \cup(2,3]\) (C) \([-4,-3) \cup[3,4)\) (D) None of these

If \(x\) is real, the maximum value of \(\frac{3 x^{2}+9 x+17}{3 x^{2}+9 x+7}\) is [2006] (A) \(1 / 4\) (B) 41 (C) 1 (D) \(17 / 7\)

Let \(\alpha\) and \(\beta\) be the roots of equation \(x^{2}-6 x-2=0\). If \(a_{n}=\alpha^{n}-\beta^{n}\), for \(n \geq 1\),then the value of \(\frac{a_{10}-2 a_{8}}{2 a_{9}}\) is equal to (A) \(-6\) (B) 3 (C) \(-3\) (D) 6

If \(2 a+3 b+6 c=0\), then at least one root of the equation \(a x^{2}+b x+c=0\) lies in the interval \(\quad\) [2004] (A) \((0,1)\) (B) \((1,2)\) (C) \((2,3)\) (D) \((1,3)\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation
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.