Problem 42
Question
For what values of \(a\) does the equation have a solution in \(x\) ? $$ a x^{2}-5=0 $$
Step-by-Step Solution
Verified Answer
Answer: The quadratic equation \(ax^2 - 5 = 0\) has real solutions for x when \(a \ge 0\).
1Step 1: Identify the constants in the equation
In the given quadratic equation, \(ax^2 - 5 = 0\), the constants are:
- a is the coefficient of the \(x^2\) term
- b is the coefficient of the x term (not given, therefore b = 0)
- c is the constant term. In this case, c = -5.
2Step 2: Calculate the discriminant
The discriminant, \(\Delta\), of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula:
$$
\Delta = b^2 - 4ac
$$
Since b = 0 in our equation, we substitute the values of a and c into the formula:
$$
\Delta = 0^2 - 4a(-5)
$$
Simplifying, we get:
$$
\Delta = 20a
$$
3Step 3: Determine the condition for real solutions
For a quadratic equation to have real solutions, the discriminant must be greater than or equal to 0. So, we will find the values of \(a\) for which \(\Delta \ge 0\).
$$
20a \ge 0
$$
Divide by 20:
$$
a \ge 0
$$
4Step 4: State the solution
The quadratic equation \(ax^2 - 5 = 0\) has real solutions for x when \(a \ge 0\).
Key Concepts
DiscriminantReal SolutionsCoefficient Identification
Discriminant
The discriminant is a key concept in understanding quadratic equations. It helps determine the nature and number of solutions an equation might have. In simple terms, the discriminant is a part of the quadratic formula that you can use to decide how many solutions a quadratic equation will have and whether they are real or complex.
For a quadratic equation in the form \[ax^2 + bx + c = 0,\]the discriminant (often represented by the symbol \( \Delta \)) is calculated using the formula:
For a quadratic equation in the form \[ax^2 + bx + c = 0,\]the discriminant (often represented by the symbol \( \Delta \)) is calculated using the formula:
- \( \Delta = b^2 - 4ac \)
- If \( \Delta > 0 \), the equation has two distinct real solutions.
- If \( \Delta = 0 \), the equation has exactly one real solution, also known as a repeated or double root.
- If \( \Delta < 0 \), the equation has two complex solutions (they are not real numbers).
Real Solutions
When dealing with quadratic equations, you will inevitably encounter the notion of real solutions. A real solution is a solution that can be represented on the number line; it is not imaginary or complex. To determine whether a quadratic equation has real solutions, the discriminant plays a critical role.
For instance, in our equation \( ax^2 - 5 = 0 \), we calculated the discriminant to be \( \Delta = 20a \). To ensure the equation has real solutions, we need this discriminant to fulfill the condition:
For instance, in our equation \( ax^2 - 5 = 0 \), we calculated the discriminant to be \( \Delta = 20a \). To ensure the equation has real solutions, we need this discriminant to fulfill the condition:
- \( \Delta \geq 0 \)
- If \( \Delta > 0 \), then two distinct real solutions exist.
- If \( \Delta = 0 \), then a single real solution exists.
Coefficient Identification
Identifying the coefficients in a quadratic equation is an initial but crucial step in solving it. Coefficients are the numbers that multiply the variables in an equation. For any quadratic equation of the form:\[ax^2 + bx + c = 0,\]understanding which terms represent \(a\), \(b\), and \(c\) is essential.
Let's break this down:
Let's break this down:
- \(a\) is the coefficient of the \(x^2\) term.
- \(b\) is the coefficient of the \(x\) term. If no \(x\) term is visible, like in our equation \(ax^2 - 5 = 0\), then \(b = 0\).
- \(c\) is the constant term, which in our problem is \(-5\).
Other exercises in this chapter
Problem 41
For what values of \(a\) does the equation have a solution in \(x\) ? $$ 2 x^{2}+a=0 $$
View solution Problem 42
Refer to Example 2 on page 379 about the value of annual gifts to Elliot growing at an annual growth factor of \(x=1+r,\) where \(r\) is the annual interest rat
View solution Problem 43
For what values of \(a\) does the equation have a solution in \(x\) ? $$ \left(a x^{2}+1\right)(x-a)=0 $$
View solution Problem 44
Refer to Example 2 on page 379 about the value of annual gifts to Elliot growing at an annual growth factor of \(x=1+r,\) where \(r\) is the annual interest rat
View solution