Problem 40 - Calculus with Concepts in Calculus

Question

Let \(f(x)=a x^{2}+b x+c\), where \(a \neq 0\). Show that any real zeros of \(f\) are given by (1). (Hint: Prove that \(a x^{2}+b x+\) \(c=0\) if and only if $$ \left(x+\frac{b}{2 a}\right)^{2}=\frac{b^{2}-4 a c}{4 a^{2}} $$ Then solve for \(x\). Note that such a (real) zero exists only if \(b^{2}-4 a c \geq 0 .\) There are two zeros if \(b^{2}-4 a c>0\), whereas there is only one zero if \(b^{2}-4 a c=0 .\) )

Step-by-Step Solution

Answer

The zeros of \(f(x)\) are given by the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 1: Recognize the Quadratic Equation

We start with the equation \( ax^2 + bx + c = 0 \). This is a standard form of a quadratic equation, and our goal is to manipulate it into a form that can be solved for its roots.

Step 2: Complete the Square

To express the quadratic equation in a different form, we complete the square:- Factor out \( a \) from the \( x^2 \) and \( x \) terms: \( a(x^2 + \frac{b}{a}x) = -c \).- Add and subtract \( \left(\frac{b}{2a}\right)^2 \) inside the parentheses: \( a((x + \frac{b}{2a})^2 - \frac{b^2}{4a^2}) = -c \).- Distribute \( a \) through: \( a(x + \frac{b}{2a})^2 - \frac{ab^2}{4a^2} = -c \).- Rearrange the equation: \( a(x + \frac{b}{2a})^2 = c + \frac{b^2}{4a} \).- Multiply through by 4 to clear denominators: \( 4a(a(x + \frac{b}{2a})^2) = 4ac + b^2 \).This simplifies to \( (x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} \).

Step 3: Analyze Discriminant

The expression \( b^2 - 4ac \) is called the discriminant. It determines the nature of the roots:- If \( b^2 - 4ac > 0 \), then \( f(x) \) has two distinct real roots.- If \( b^2 - 4ac = 0 \), then \( f(x) \) has exactly one real root (a repeated root).- If \( b^2 - 4ac < 0 \), then \( f(x) \) has no real roots.

Step 4: Solve for Real Roots

Assume \( b^2 - 4ac \geq 0 \):- Take the square root of both sides of \( (x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} \).- This gives two equations: \( x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \).- Solving these, we find \( x = -\frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a} \) and \( x = -\frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a} \).

Step 5: Present the Formula for Real Zeros

The solutions for \( x \), also known as the roots of the quadratic equation, can be expressed as:\[x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a}\]These are the points where the quadratic function crosses the x-axis, provided that the discriminant is non-negative.

Key Concepts

Completing the Square

Completing the square is a method used to solve quadratic equations by transforming them into a perfect square trinomial. This technique helps us rewrite the equation in a way that makes it easier to solve for the roots of the quadratic.

• Start with a quadratic equation in the form: \( ax^2 + bx + c = 0 \).
• Factor out the coefficient of the quadratic term, \( a \), from the first two terms: \( a(x^2 + \frac{b}{a}x) = -c \).
• Add and subtract the square of half the coefficient of \( x \) within the parenthesis: \( a((x + \frac{b}{2a})^2 - \frac{b^2}{4a^2}) = -c \).
• Rearrange to form a perfect square trinomial: \( a(x + \frac{b}{2a})^2 = c + \frac{b^2}{4a} \).
• Solve the resulting equation: \( (x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} \).

By completing the square, we convert a difficult equation into one that is much simpler to solve. This lays the foundation for understanding the solutions in terms of the vertex form of the quadratic.

Discriminant

The discriminant of a quadratic equation is a key component in determining the type and number of solutions the equation will have. It is represented by the expression \( b^2 - 4ac \) taken from the quadratic formula.Let's delve into what the discriminant can tell us:

• When \( b^2 - 4ac > 0 \), the quadratic equation has two distinct real roots. This means the parabola representing the quadratic function crosses the x-axis at two different points.
• If \( b^2 - 4ac = 0 \), the quadratic has exactly one real root, known as a repeated or double root. Here, the parabola touches the x-axis at only one point.
• Lastly, when \( b^2 - 4ac < 0 \), there are no real roots. The parabola does not intersect the x-axis, indicating that any solutions would be complex rather than real.

The discriminant provides a quick way to predict the number and nature of the roots without solving the equation.

Roots of a Quadratic

The roots of a quadratic equation are the values of \( x \) for which the equation \( ax^2 + bx + c = 0 \) holds true. These are the points where the quadratic curve intersects the x-axis. Solving for the roots is a fundamental aspect of working with quadratics and can be done using the quadratic formula derived through completing the square.To find the roots:

• Begin by setting the equation in the form \( ax^2 + bx + c = 0 \).
• Utilize the quadratic formula: \[x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a}\]
• This formula gives you two solutions: one using the plus sign and the other with the minus sign preceding the square root term. These solutions are the roots of the quadratic.

The roots can be real or complex, depending on the value of the discriminant. If the discriminant is greater than or equal to zero, the solutions are real, bringing us to the next concept - real zeros.

Real Zeros

Real zeros are the points where the quadratic function \( f(x) = ax^2 + bx + c \) crosses the x-axis. They are also the solutions to the quadratic equation \( ax^2 + bx + c = 0 \). Determining whether zeros are real depends on the value of the discriminant \( b^2 - 4ac \).For real zeros:

• When \( b^2 - 4ac \geq 0 \), real zeros exist. The quadratic function will intersect the x-axis at either one or two points.
• If \( b^2 - 4ac > 0 \), there are two distinct real zeros, showing that the quadratic curve crosses the x-axis twice.
• In the case where \( b^2 - 4ac = 0 \), there is exactly one real zero. Here, the curve just touches the x-axis at a single point, known as a vertex.

Real zeros are crucial in understanding the real-world behaviors modeled by quadratic functions, as they often represent solutions to practical problems.