Chapter 1: Problem 40
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 \(b^{2}-4 a c=0 .\) )
Short Answer
Step by step solution
Recognize the Quadratic Equation
Complete the Square
Analyze Discriminant
Solve for Real Roots
Present the Formula for Real Zeros
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
- 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: \( (x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2} \).
Discriminant
- 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.
Roots of a Quadratic
- 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.
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.