Question

Suppose the quadratic equation \(a x^2+5 x+c=0\) has one real solution. Is it possible for \(a\) and \(c\) to be integers? rational numbers? Explain your reasoning. Then describe the possible values of \(a\) and \(c\).

Short answer

It is not possible for \(a\) and \(c\) to be integers, but it is possible for \(a\) and \(c\) to be rational numbers for the quadratic equation \( ax^2+5x+c=0\) to only have one real solution.

Step-by-step solution

Setup the Discriminant

First, identify \(b^2-4ac\) – the discriminant, of our quadratic equation. Since \(a = a, b= 5, c = c\), the discriminant becomes \(5^2-4ac\). Set it equal to zero, this is because we're told the quadratic has only one real solution.

Solve for ac

We solve the discriminant equals to zero giving us: \(5^2-4ac = 0 \). Resolving for \(ac\) we get: \(ac = 5^2/4 = 6.25\).

Identify the nature of a and c

Since \(ac=6.25\), and \(ac\) is a product of \(a\) and \(c\), we know that \(a\) and \(c\) can not be integers, as no two integers multiply to give a non-integer. However, \(a\) and \(c\) can be rational numbers, as \(6.25\) can be expressed as a product of two rational numbers in multiple ways. For example, \(a =2.5\) and \(c = 2.5\) or \(a =1.25\) and \(c =5\), among others.

Key concepts

Understanding the Discriminant

The discriminant is a component of the quadratic formula that helps determine the nature of the roots of a quadratic equation. In any quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is given by \( b^2 - 4ac \). This value is crucial in telling us about the solutions of the equation.

• If the discriminant is positive, \( \Delta > 0 \), there are two distinct real solutions.
• If the discriminant is zero, \( \Delta = 0 \), there is exactly one real solution, which is also called a repeated or double root.
• If the discriminant is negative, \( \Delta < 0 \), there are no real solutions, and instead, we have two complex solutions.

In our specific equation \( ax^2 + 5x + c = 0 \), we have a discriminant of \( 5^2 - 4ac \). For the scenario with one real solution, setting this equal to zero ensures our discriminant condition is met.

Identifying Real Solutions

When a quadratic equation has one real solution, it means that the graph of the equation is tangent to the x-axis at a single point. This situation occurs only when the discriminant is zero. Therefore, for our equation \( ax^2 + 5x + c = 0 \), the condition for having one real solution is that the discriminant \( 5^2 - 4ac \) equals zero.Solving \( 25 - 4ac = 0 \), we find that \( ac = 6.25 \). This tells us that for the given quadratic equation to have one real solution, the product of coefficients \( a \) and \( c \) must equal 6.25. Thus, given the constraints where \( a \) and \( c \) cannot be integers, they must be organized as rational numbers.

Rational Numbers in Quadratics

Rational numbers are numbers that can be expressed as fractions \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q eq 0 \). In the context of our quadratic equation, if \( ac = 6.25 \), \( a \) and \( c \) need to be rational.This is because 6.25 can be written in many ways using rational numbers, such as \( \frac{25}{4} \). Using this, possible combinations for \( a \) and \( c \) can be \( a = 2.5 \), \( c = 2.5 \) or \( a = 1.25 \), \( c = 5 \). Each pair of rational numbers will satisfy \( ac = 6.25 \) without being integers.

Integer Solutions and Their Impossibility

Though integer solutions can often solve quadratic equations, in this case, having \( ac = 6.25 \) implies that integers are not possible solutions for \( a \) and \( c \).Integer solutions would mean \( a \times c \) is a whole number. However, 6.25 cannot be formed by multiplying two whole numbers. Integer multiplication only results in whole numbers, and since 6.25 is not a whole number, neither \( a \) nor \( c \) can be integers.Thus, solving our given quadratic with one real solution mandates exploring beyond integers to rational numbers.