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(a) Let \(a, b,\) and \(c\) be real numbers with \(a \neq 0 .\) Explain how to use a part of the quadratic formula (called the discriminant) to determine if the quadratic equation \(a x^{2}+b x+c=0\) has two real number solutions, one real number solution, or no real number solutions. (See Exercise ( 11 ) in Section 1.2 for a statement of the quadratic formula.) (b) Prove that if \(a, b,\) and \(c\) are real numbers for which \(a>0\) and \(c<0\), then one solution of the quadratic equation \(a x^{2}+b x+c=0\) is a positive real number. (c) Prove that if \(a, b,\) and \(c\) are real numbers, if \(a \neq 0, b>0\) and \(\frac{b}{2}<\sqrt{a c},\) then the quadratic equation \(a x^{2}+b x+c=0\) has no real number solution.

Short Answer

Expert verified
(a) The discriminant is \(b^2 - 4ac\). There are two distinct real number solutions if the discriminant is positive, one real number solution if the discriminant is zero, and no real number solutions if the discriminant is negative. (b) Given \(a > 0\) and \(c < 0\), the function \(f(x) = ax^2 + bx + c\) is an upward-opening parabola with a negative value at \(x=0\). By the Intermediate Value Theorem, there must be at least one positive real number solution. (c) Given \(a \neq 0\), \(b > 0\), and \(\frac{b}{2}<\sqrt{ac}\), squaring both sides yields \(\frac{b^2}{4}<ac\) or \(b^2 < 4ac\), implying a negative discriminant. Thus, there are no real number solutions for the quadratic equation \(ax^2 + bx + c = 0\).

Step by step solution

01

State the Quadratic formula

The quadratic formula is used to find the solutions of a quadratic equation \(ax^2 + bx + c = 0\), and it is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
02

Explain the discriminant

The discriminant is the part inside the square root in the quadratic formula, i.e., \(b^2 - 4ac\). It is used to determine the nature of the solutions of a quadratic equation: 1. If the discriminant is positive (\(b^2 - 4ac > 0\)), there are two distinct real number solutions. 2. If the discriminant is zero (\(b^2 - 4ac = 0\)), there is exactly one real number solution (a repeated root). 3. If the discriminant is negative (\(b^2 - 4ac < 0\)), there are no real number solutions (the solutions are complex). (b)
03

Given conditions

We are given that \(a > 0\) and \(c < 0\), and we need to prove that one solution of the quadratic equation \(ax^2 + bx + c = 0\) is a positive real number.
04

Analyze the quadratic function

We can rewrite the given equation as \(f(x) = ax^2 + bx + c\). Since \(a > 0\), the function represents an upward-opening parabola. As \(c < 0\), the value of the function at \(x=0\) is negative (i.e., \(f(0) = c < 0\)).
05

Show the existence of a positive solution

As \(f(x)\) is continuous and its graph represents an upward-opening parabola, there must exist an as-yet-undetermined positive value of x where \(f(x) > 0\). Therefore, by the Intermediate Value Theorem, at least one solution of the equation \(ax^2 + bx + c = 0\) must be a positive real number. (c)
06

Given conditions

We are given that \(a \neq 0\), \(b > 0\), and \(\frac{b}{2}<\sqrt{ac}\), and we need to prove that the quadratic equation \(ax^2 + bx + c = 0\) has no real number solution.
07

Rewrite the given inequality

We are given that \(\frac{b}{2}<\sqrt{ac}\). Squaring both sides, we have: \[\frac{b^2}{4}<ac\]
08

Discriminant analysis

To prove that there are no real number solutions, we need to show that the discriminant is negative. The discriminant is given by \(b^2 - 4ac\). We already have \(\frac{b^2}{4}<ac\), which can be written as: \[b^2<4ac\] Thus, \(b^2 - 4ac < 0\). Therefore, the discriminant is negative, and the quadratic equation \(ax^2 + bx + c = 0\) has no real number solutions.

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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 that helps us understand the nature of the solutions of a quadratic equation. In the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the expression \(b^2 - 4ac\) is known as the discriminant. It can tell us how many real number solutions, or roots, a quadratic equation has. Here's how:
  • If the discriminant is greater than zero \( (b^2 - 4ac > 0) \), the quadratic equation has two distinct real number solutions. This means the graph of the quadratic will intersect the x-axis at two points.
  • If the discriminant is exactly zero \( (b^2 - 4ac = 0) \), the equation has exactly one real solution, sometimes called a repeated or double root. In this situation, the graph of the quadratic touches the x-axis at precisely one point.
  • Finally, if the discriminant is less than zero \( (b^2 - 4ac < 0) \), there are no real number solutions. This occurs when the graph of the quadratic does not intersect the x-axis at all, resulting in complex or imaginary solutions instead.
Understanding the discriminant is essential for predicting the behavior of quadratic equations.
Real Number Solutions
Real number solutions to quadratic equations are determined by the discriminant. When solving a quadratic equation of the form \(ax^2 + bx + c = 0\), the goal is to find the values of \(x\) that fulfill this equation. These values are called roots or solutions, and they can be real or complex numbers depending on the discriminant.For those looking for real number solutions:
  • When the discriminant \((b^2 - 4ac)\) is positive, it indicates that the quadratic equation has two distinct real solutions. The roots differ from each other and show the points where the parabola crosses the x-axis.
  • With a zero discriminant, the quadratic equation has one real solution, known as a double root. Here, the parabola just touches the x-axis at one point, meaning that both roots are the same.
If the discriminant is negative, the result would be complex solutions, which are not real numbers. Knowing how to find and interpret these real number solutions is key to understanding quadratic equations.
Intermediate Value Theorem
The Intermediate Value Theorem is a mathematical concept that can be applied to continuous functions, such as quadratic functions. It states that if a function is continuous on a closed interval \([a, b]\), and if \(f(a)\) and \(f(b)\) are of opposite signs, then there is at least one \(c\) in the interval \([a, b]\) such that \(f(c) = 0\).In the context of quadratic equations, the theorem helps to prove the existence of a real root between two points. Suppose you have a quadratic equation \(f(x) = ax^2 + bx + c\) where \(a > 0\) and \(c < 0\). If \(f(0) = c < 0\), indicating the function's value is negative at \(x = 0\), and if the parabola opens upwards, then the function must eventually reach a positive value for some positive \(x\).Applying the Intermediate Value Theorem, we deduce there must be at least one value of \(x\) in the positive range where \(f(x) = 0\), hence indicating a real number solution.
Upward-Opening Parabola
An upward-opening parabola is the graph of a quadratic function where the coefficient of \(x^2\), known as \(a\), is positive \( (a > 0) \). This means the parabola curves upwards like a U-shape.Key features of an upward-opening parabola include:
  • The vertex represents the lowest point, also known as the minimum point of the parabola.
  • The parabola is symmetric about a vertical line through the vertex, which is called the axis of symmetry.
  • If the discriminant is positive, the parabola crosses the x-axis at two points, indicating two real roots.
  • If the discriminant is zero, the vertex will touch the x-axis, indicating one real root.
  • If the discriminant is negative, the parabola lies entirely above the x-axis, meaning there are no real roots.
Understanding the direction an upward-opening parabola faces is crucial when analyzing quadratic functions, especially when predicting the number of real solutions.

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Most popular questions from this chapter

We have already seen examples of Pythagorean triples, which are natural numbers \(a, b,\) and \(c\) where \(a^{2}+\) \(b^{2}=c^{2}\). For example, \(3,4,\) and 5 form a Pythagorean triple as do \(5,\) \(12,\) and \(13 .\) One of the famous mathematicians of the 17 th century was Pierre de Fermat \((1601-1665)\). Fermat made an assertion that for each natural number \(n\) with \(n \geq 3,\) there are no positive integers \(a, b,\) and \(c\) for which \(a^{n}+b^{n}=c^{n}\). This assertion was discovered in a margin of one of Fermat's books after his death, but Fermat provided no proof. He did, however, state that he had discovered a truly remarkable proof but the margin did not contain enough room for the proof. This assertion became known as Fermat's Last Theorem but it more properly should have been called Fermat's Last Conjecture. Despite the efforts of mathematicians, this "theorem" remained unproved until Andrew Wiles, a British mathematician, first announced a proof in June of \(1993 .\) However, it was soon recognized that this proof had a serious gap, but a widely accepted version of the proof was published by Wiles in \(1995 .\) Wiles' proof uses many concepts and techniques that were unknown at the time of Fermat. We cannot discuss the proof here, but we will explore and prove the following proposition, which is a (very) special case of Fermat's Last Theorem. Proposition. There do not exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\) Although Fermat's Last Theorem implies this proposition is true, we will use a proof by contradiction to prove this proposition. For a proof by contradiction, we assume that there exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\). Since 2 is the only even prime number, we will use the following cases: (1) \(a=b=2 ;\) (2) \(a\) and \(b\) are both odd; and (3) one of \(a\) and \(b\) is odd and the other one is 2 . (a) Show that the case where \(a=b=2\) leads to a contradiction and hence, this case is not possible. (b) Show that the case where \(a\) and \(b\) are both odd leads to a contradiction and hence, this case is not possible. (c) We now know that one of \(a\) or \(b\) must be equal to 2 . So we assume that \(b=2\) and that \(a\) is an odd prime. Substitute \(b=2\) into the equation \(b^{3}=c^{3}-a^{3}\) and then factor the expression \(c^{3}-a^{3}\). Use this to obtain a contradiction. (d) Write a complete proof of the proposition.

(Exercise (18), Section 3.2) Prove the following proposition: Let \(a\) and \(b\) be integers with \(a \neq 0\). If \(a\) does not divide \(b\), then the equation \(a x^{3}+b x+(b+a)=0\) does not have a solution that is a natural number.

For a right triangle, suppose that the hypotenuse has length \(c\) feet and the lengths of the sides are \(a\) feet and \(b\) feet. (a) What is a formula for the area of this right triangle? What is an isosceles triangle? (b) State the Pythagorean Theorem for right triangles. \(\star\) (c) Prove that the right triangle described above is an isosceles triangle if and only if the area of the right triangle is \(\frac{1}{4} c^{2}\).

Prove that for each integer \(a\), if \(a^{2}-1\) is even, then 4 divides \(a^{2}-1\).

Congruence Modulo 6 . (a) Find several integers that are congruent to 5 modulo 6 and then square each of these integers. (b) For each integer \(m\) from Part (20a), determine an integer \(k\) so that \(0 \leq k<6\) and \(m^{2} \equiv k(\bmod 6) .\) What do you observe? (c) Based on the work in Part ( \(20 \mathrm{~b}\) ), complete the following conjecture: For each integer \(m,\) if \(m \equiv 5(\bmod 6),\) then \(\ldots\) (d) Complete a know-show table for the conjecture in Part (20c) or write a proof of the conjecture.

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