Problem 80
Question
Explain why the quadratic formula gives solutions only if \(a \neq 0\) and \(b^{2}-4 a c \geq 0\).
Step-by-Step Solution
Verified Answer
The quadratic formula gives real solutions only if \(a \neq 0\) to avoid division by zero and retain the quadratic nature of the equation, and \(b^2 - 4ac \geq 0\) to ensure the square root, or discriminant, is real.
1Step 1: Understand the Quadratic Formula
The quadratic formula is used to solve the quadratic equation of the form \(ax^2 + bx + c = 0\). The solutions of this equation are given by the quadratic formula as: \(x = \[-b \pm \sqrt{b^2 - 4ac}\] / 2a\). The question asks about the conditions \(a \neq 0\) and \(b^2 - 4ac \geq 0\)
2Step 2: Analyze the condition \(a \neq 0\)
The coefficient \(a\) of \(x^2\) is in the denominator of the quadratic formula. Mathematically, division by zero is undefined. Thus, \(a\) cannot be zero for this formula to make sense. If \(a = 0\), the equation would not be a quadratic one, but a linear equation instead.
3Step 3: Analyze the condition \(b^2 - 4ac \geq 0\)
The term under the square root in the quadratic formula, \(b^2 - 4ac\), is called the discriminant. It determines the nature of the roots. If the discriminant is negative, the square root of a negative number becomes an imaginary number. Thus, for real-valued solutions, the discriminant has to be greater than or equal to zero, i.e \(b^2 - 4ac \geq 0\) is required.
Key Concepts
Quadratic EquationDiscriminantReal-Valued Solutions
Quadratic Equation
A quadratic equation is a type of polynomial equation of the second degree. In its standard form, it is written as \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\) because if \(a\) were zero, the equation would become linear rather than quadratic. The highest power of the variable \(x\) in this equation is 2, which is why it is called a quadratic equation. Quadratic equations frequently appear in various fields, such as physics, engineering, and economics, due to their ability to model parabolic shapes and phenomena that follow a parabolic trajectory. Key points to remember about quadratic equations:
- The standard form is \(ax^2 + bx + c = 0\).
- Coefficient \(a\) cannot be zero.
- The quadratic equation can have two, one, or no real solutions depending on the discriminant.
Discriminant
The discriminant is a crucial part of the quadratic formula, represented by the expression \(b^2 - 4ac\). It helps determine the nature of the solutions of the quadratic equation. The discriminant uncovers what kind of roots the equation might have based on its value:
- If \(b^2 - 4ac > 0\), the equation has two distinct real solutions.
- If \(b^2 - 4ac = 0\), the equation has exactly one real solution, or a repeated real root.
- If \(b^2 - 4ac < 0\), the equation has no real solutions; instead, it has two complex or imaginary solutions.
Real-Valued Solutions
Real-valued solutions are the solutions to an equation that do not involve imaginary numbers. For a quadratic equation, real-valued solutions occur when the discriminant \(b^2 - 4ac\) is non-negative. Specifically:
- If the discriminant is zero, it means the parabola touches the x-axis at only one point. Thus, the quadratic equation has exactly one real solution, indicated by a repeated root.
- If the discriminant is positive, the parabola intersects the x-axis at two separate points, meaning the quadratic equation has two distinct real solutions.
Other exercises in this chapter
Problem 79
VERTEX FORM The vertex form of a quadratic function is \(y=a(x-h)^{2}+k\). Its graph is a parabola with vertex at \((\boldsymbol{h}, \boldsymbol{k})\). Use comp
View solution Problem 80
Find the product. $$(x+4)(3 x-7)$$
View solution Problem 81
Using \(4 \sqrt{x}=2 x+k,\) find three different expressions that can be substituted for \(k\) so that the equation has two solutions, one solution, and no solu
View solution Problem 81
Find the product. $$(2 x-3)(5 x-9)$$
View solution