Problem 10
Question
In an exercise in Section 3.2 we proved that the quadratic equation \(a x^{2}+b x+c=0\) has two solutions if \(a c<0 .\) Find a counterexample which shows that this implication cannot be replaced with a biconditional.
Step-by-Step Solution
Verified Answer
The equation \( -x^2 + 3x - 2 = 0 \) shows a*c > 0 and still has two solutions.
1Step 1 - Understand the Statement
The given condition states that the quadratic equation has two solutions if and only if (iff) the product of 'a' and 'c' is less than 0. A counterexample will show a case where the implication holds but the reverse (iff) does not.
2Step 2 - Recall the Quadratic Formula
The solutions for the quadratic equation are given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. The discriminant, \[ b^2 - 4ac \], determines the nature of the roots. The equation has two real solutions if the discriminant is greater than 0.
3Step 3 - Analyze the Discriminant
For the equation to have two solutions, we need \[ b^2 - 4ac > 0 \], regardless of the sign of a and c.
4Step 4 - Construct the Counterexample
Choose a quadratic equation that has two solutions, but with a*c > 0. For example, consider the equation \( x^2 - 2x + 1 = 0 \). Here, \( a = 1 \), \( b = -2 \), and \( c = 1 \). The product a*c = 1 which is greater than 0.
5Step 5 - Verify the Equation's Solutions
Check that the discriminant is zero: \[ b^2 - 4ac = (-2)^2 - 4(1)(1) = 4 - 4 = 0 \]. The equation has a double root, not two distinct solutions. A better counterexample is \( -x^2 + 3x - 2 \), where \( a=-1 \), \( b=3 \), \( c=-2 \). Here, a * c > 0 but the discriminant is positive.
6Step 6 - Confirm the Counterexample
For \( -x^2 + 3x - 2 \): \( a = -1 \), \( c = -2 \) and so a * c = 2 > 0. The discriminant is \( 3^2 - 4(-1)(-2) = 9 - 8 = 1 \), so the equation has two solutions.
Key Concepts
DiscriminantBiconditional StatementsRoots of Polynomial Equations
Discriminant
In quadratic equations, the discriminant is a key element that helps determine the nature of the solutions. It is given by the formula: \ \ \[ b^2 - 4ac \]. This value can tell us if the quadratic equation has two distinct real roots, a single real root, or no real roots.
\*If the discriminant is greater than 0, the equation has two distinct real solutions.
\*If the discriminant is equal to 0, there is exactly one real solution, which is also called a double root.
\*If the discriminant is less than 0, there are no real solutions, only complex solutions.
In the solution, the discriminant is used to prove the presence of two solutions, focusing on the conditions that make it greater than 0.
\*If the discriminant is greater than 0, the equation has two distinct real solutions.
\*If the discriminant is equal to 0, there is exactly one real solution, which is also called a double root.
\*If the discriminant is less than 0, there are no real solutions, only complex solutions.
In the solution, the discriminant is used to prove the presence of two solutions, focusing on the conditions that make it greater than 0.
Biconditional Statements
Biconditional statements in mathematics assert that two conditions are both necessary and sufficient for each other. This is typically written as 'if and only if' (iff).
In this exercise, we are asked to see if the condition where a quadratic equation has two solutions 'if and only if' the product of 'a' and 'c' is less than 0 is true.
However, using the counterexample provided, we see that a quadratic equation can have two solutions even if a * c > 0. This means that the 'if and only if' statement does not hold true universally.
Thus, while 'a * c < 0' is a sufficient condition for the quadratic equation to have two solutions, it is not a necessary one.
In this exercise, we are asked to see if the condition where a quadratic equation has two solutions 'if and only if' the product of 'a' and 'c' is less than 0 is true.
However, using the counterexample provided, we see that a quadratic equation can have two solutions even if a * c > 0. This means that the 'if and only if' statement does not hold true universally.
Thus, while 'a * c < 0' is a sufficient condition for the quadratic equation to have two solutions, it is not a necessary one.
Roots of Polynomial Equations
Roots are solutions to equations; for polynomial equations, these roots are the values of the variable that satisfy the equation.
For second-degree polynomial equations (quadratic equations), there can be two roots. These roots can be real and distinct, real and the same (double roots), or complex.
Using the quadratic formula \ \ \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], we solve for the roots of the quadratic equation. Understanding how to manipulate this formula and the discriminant helps in determining the nature of these roots.
The exercise provides a clear method to demonstrate that even if the product of 'a' and 'c' is greater than 0 (a * c > 0), the quadratic equation can still have two distinct real roots. This explains why 'a * c < 0' as the sole condition for two solutions is not universally applicable.
For second-degree polynomial equations (quadratic equations), there can be two roots. These roots can be real and distinct, real and the same (double roots), or complex.
Using the quadratic formula \ \ \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], we solve for the roots of the quadratic equation. Understanding how to manipulate this formula and the discriminant helps in determining the nature of these roots.
The exercise provides a clear method to demonstrate that even if the product of 'a' and 'c' is greater than 0 (a * c > 0), the quadratic equation can still have two distinct real roots. This explains why 'a * c < 0' as the sole condition for two solutions is not universally applicable.
Other exercises in this chapter
Problem 9
True or false: Whenever an integer \(n\) is a divisor of the square of an integer, \(m^{2},\) it follows that \(n\) is a divisor of \(m\) as well. (In symbols,
View solution Problem 9
Prove that \(\forall x \in \mathbb{R}, x \notin \mathbb{Z} \Longrightarrow\lfloor x\rfloor+\lfloor-x\rfloor=-1\).
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Suppose that \(a, b\) and \(c\) are integers such that \(a \mid b\) and \(b \mid c .\) Prove that \(a \mid c\)
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Suppose that \(a, b, c\) and \(d\) are integers with \(a \neq c .\) Further, suppose that \(x\) is a real number satisfying the equation $$ \frac{a x+b}{c x+d}=
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