Problem 81
Question
For each pair of numbers, find the values of \(a, b,\) and \(c\) for which the quadratic equation ax \(^{2}+b x+c=0\) has the given numbers as solutions. Answers may vary. (Hint: Use the zero-product property in reverses $$2 i,-2 i$$
Step-by-Step Solution
Verified Answer
The values are \(a = 1\), \(b = 0\), \(c = 4\).
1Step 1: Identify the solutions
The problem provides the solutions to the quadratic equation as \(2i\) and \(-2i\). In a quadratic equation, these are the roots.
2Step 2: Set up factored form
Since \(2i\) and \(-2i\) are the roots of the quadratic, we can write the equation in factored form: \((x - 2i)(x + 2i) = 0\).
3Step 3: Apply zero-product property
According to the zero-product property, if the product of two expressions is zero, then at least one of the expressions must be zero. In our case, \((x - 2i) = 0\) or \((x + 2i) = 0\) will give us the roots \(2i\) and \(-2i\).
4Step 4: Expand the factored form
Next, we expand \((x - 2i)(x + 2i) = 0\) using the difference of squares formula:\[ (x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 - (4)(-1) = x^2 + 4 \]
5Step 5: Identify values of \(a\), \(b\), and \(c\)
Now that we have \(x^2 + 4 = 0\), we can see that in the standard form \(ax^2 + bx + c = 0\), the coefficients are:- \(a = 1\)- \(b = 0\)- \(c = 4\)
Key Concepts
Zero-Product PropertyComplex RootsDifference of Squares
Zero-Product Property
The zero-product property is a key principle used in solving quadratic equations, among other algebraic equations. It states that if the product of two or more factors is zero, then at least one of the factors must be zero. This property is fundamental when dealing with equations set to zero because it allows us to solve for the roots or solutions of the equation by setting each factor equal to zero.
For instance, in the problem, we have
These equations are easy to solve:
For instance, in the problem, we have
(x - 2i)(x + 2i) = 0.
(x - 2i) = 0 or (x + 2i) = 0.
These equations are easy to solve:
- First equation yields the root
x = 2i - Second equation gives
x = -2i
Complex Roots
Complex roots arise when the solutions to a polynomial equation, like a quadratic, involve imaginary numbers. An imaginary number is a multiple of 'i', where 'i' is the square root of -1.
In the exercise, we're dealing with the complex roots
Complex roots often appear in pairs when the quadratic equation's discriminant is negative. These pairs, consisting of one root with
Understanding complex roots is crucial for solving and interpreting equations that stretch beyond the realm of real numbers, opening up a wider field of possibilities in problem-solving applications.
In the exercise, we're dealing with the complex roots
2iand-2i
Complex roots often appear in pairs when the quadratic equation's discriminant is negative. These pairs, consisting of one root with
+i and another with -i, result in no real number solutions for the equation, but instead ones involving imaginary units.
Understanding complex roots is crucial for solving and interpreting equations that stretch beyond the realm of real numbers, opening up a wider field of possibilities in problem-solving applications.
Difference of Squares
The difference of squares is a fundamental algebraic identity used to factor expressions such as
In the given problem, the factored form
(x^2 - y^2).
x^2 - y^2 = (x - y)(x + y).
In the given problem, the factored form
(x - 2i)(x + 2i)
x^2 - (2i)^2
x^2 - (-4)which simplifies tox^2 + 4.
Other exercises in this chapter
Problem 80
Find the center-radius form of the circle with the given equation. Determine the coordinates of the center, find the radius, and graph the circle. $$x^{2}+y^{2}
View solution Problem 80
Simplify each expression to \(i, 1,-i,\) or \(-1\) $$i^{-15}$$
View solution Problem 81
Find the center-radius form of the circle with the given equation. Determine the coordinates of the center, find the radius, and graph the circle. $$x^{2}+y^{2}
View solution Problem 81
Simplify each expression to \(i, 1,-i,\) or \(-1\) $$\frac{1}{i^{9}}$$
View solution