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Chapter 1: Problem 81

Show that the function \(f(x)=x^{2}+b x-1\) has a real zero for any value \(b\)

Short Answer

Expert verified
Given the quadratic equation \(f(x) = x^{2} + bx - 1\), it can be observed that the discriminant (\(b^{2}-4ac\)) is always positive for any real \(b\), hence there will always be a real zero.

Step by step solution

01

Identify coefficients for the quadratic formula

In this equation, \(f(x) = x^{2} + bx - 1\), the coefficients are \(a = 1\), \(b = b\) as given, and \(c = -1\).
02

Substitute coefficients into the quadratic formula

Substituting these values into the quadratic formula, we get \(x=\frac{-b卤\sqrt{b^{2}-4(1)(-1)}}{2(1)}\), which simplifies to \(x=\frac{-b卤\sqrt{b^{2}+4}}{2}\).
03

Consider the discriminant

The importance of the quadratic formula is that it allows us to see that a solution exists if the discrimant (\(b^{2}-4ac\)) is positive. In our case, it is \(b^{2}+4\), which is always positive for any real \(b\). Thus, there is always a real zero.

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Key Concepts

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

Discriminant
The discriminant is a crucial part of the quadratic formula, providing insight into the nature of the solutions for a quadratic equation. For a quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant is given by:
  • \(D = b^2 - 4ac\)
This value tells us several things about the roots of the equation:
  • If \(D > 0\), there are two distinct real solutions.
  • If \(D = 0\), there is exactly one real solution, also known as a repeated or double root.
  • If \(D < 0\), the equation has no real solutions but two complex solutions instead.
For the quadratic function \(f(x)=x^2 + bx - 1\), the discriminant becomes \(b^2 + 4\), which is always positive for any real \(b\). This guarantees at least one real root.
Real Solutions
Real solutions are the intersection points of the function with the x-axis. A quadratic equation may have one, two, or no real solutions, which can be determined using the discriminant. For equations like \(x^2 + bx - 1 = 0\), we need to examine
  • The sign and value of the discriminant.
  • The value of \(D = b^2 + 4\) in our scenario, which is always positive.
Since \(b^2 + 4 > 0\) regardless of the specific value of \(b\), there are always two real solutions. This means the graph of \(f(x)\) will cross the x-axis at two points.
Quadratic Formula
The quadratic formula provides a method to find the solutions of any quadratic equation \(ax^2 + bx + c = 0\). The formula is written as:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
It is derived from completing the square of the quadratic expression. By substituting the coefficients of the given equation into the formula, we can find its roots. In the case of \(f(x) = x^2 + bx - 1\), substituting gives:
  • \(x = \frac{-b \pm \sqrt{b^2 + 4}}{2}\)
This solution is valid for any real value of \(b\), thus showing that the quadratic has real solutions for any given \(b\).
Coefficients
Coefficients are the numerical values placed in front of the variables in a polynomial equation. They play a fundamental role in defining the equation and figuring out its behavior. In our quadratic equation \(f(x)=x^2 + bx - 1\), there are:
  • \(a = 1\), which is the coefficient of \(x^2\).
  • \(b = b\), which varies and is the coefficient of \(x\).
  • \(c = -1\), which is the constant term and does not multiply with any variable.
These coefficients are crucial for using the quadratic formula. When substituted into the formula, they reveal whether the function has real solutions by determining the sign of the discriminant.

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Most popular questions from this chapter

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