If c > 0 and 4a + c < 2b, then ax 2– bx + c = 0 has a root in which one of the following intervals?
(0, 2)
Finding the Root Interval for the Quadratic Equation
We are given a quadratic equation ${ax^2 - bx + c = 0}$ and two conditions: ${c > 0}$ and ${4a + c < 2b}$. We need to determine which of the given intervals contains a root of this equation.
Let the function be ${f(x) = ax^2 - bx + c}$. A root of the equation ${ax^2 - bx + c = 0}$ is a value of ${x}$ for which ${f(x) = 0}$. If a continuous function changes sign between two points, say ${x_1}$ and ${x_2}$, then there must be at least one root between ${x_1}$ and ${x_2}$. Quadratic functions are continuous.
Evaluating the Function at Key Points
Let's evaluate the function ${f(x)}$ at ${x=0}$ and ${x=2}$, as these points relate to the structure of the given inequality ${4a + c < 2b}$ and the first interval option:
- At ${x = 0}$:
${f(0) = a(0)^2 - b(0) + c = c}$
We are given the condition ${c > 0}$. Therefore, ${f(0) > 0}$. - At ${x = 2}$:
${f(2) = a(2)^2 - b(2) + c = 4a - 2b + c}$
We are given the condition ${4a + c < 2b}$. Let's rearrange this inequality:${4a + c - 2b < 0}$
Comparing this to ${f(2) = 4a - 2b + c}$, we can see that ${f(2) = 4a + c - 2b}$. Therefore, based on the given inequality, ${f(2) < 0}$.
Determining the Interval Containing a Root
We have found that:
- ${f(0) > 0}$ (the function is positive at ${x=0}$)
- ${f(2) < 0}$ (the function is negative at ${x=2}$)
Since the function ${f(x)}$ is a continuous function (it's a quadratic), and its value changes from positive at ${x=0}$ to negative at ${x=2}$, it must cross the x-axis at least once between ${x=0}$ and ${x=2}$. The point where it crosses the x-axis is a root of the equation ${f(x) = 0}$.
This means there is a root in the open interval ${ (0, 2) }$.
This conclusion is based on the Intermediate Value Theorem, which states that for a continuous function on a closed interval $[a, b]$, if $f(a)$ and $f(b)$ have opposite signs, then there exists at least one value $c$ in the open interval $(a, b)$ such that $f(c) = 0$. In our case, the interval is $(0, 2)$.
Verification with Other Intervals
While we have found a root in $(0, 2)$, let's quickly consider if the conditions guarantee roots in other intervals. The given conditions ${c > 0}$ and ${4a + c < 2b}$ directly lead to ${f(0) > 0}$ and ${f(2) < 0}$. This sign change guarantees a root between 0 and 2. The other intervals do not necessarily show a sign change based *only* on the given conditions.
| Interval | Endpoints | Function Values | Sign Change? | Root Guaranteed? |
|---|---|---|---|---|
| (0, 2) | x=0, x=2 | f(0) > 0, f(2) < 0 | Yes | Yes |
| (2, 3) | x=2, x=3 | f(2) < 0, f(3) = 9a - 3b + c | Not guaranteed by conditions | No |
| (3, 4) | x=3, x=4 | f(3) = 9a - 3b + c, f(4) = 16a - 4b + c | Not guaranteed by conditions | No |
| (-2, 0) | x=-2, x=0 | f(-2) = 4a + 2b + c, f(0) > 0 | Not guaranteed by conditions | No |
Based on our analysis, the only interval guaranteed to contain a root by the given conditions is ${ (0, 2) }$.
Revision Table: Quadratic Equation Roots and Intervals
| Concept | Description | Relevance to Problem |
|---|---|---|
| Quadratic Equation | An equation of the form ${ax^2 + bx + c = 0}$, where ${a \neq 0}$. | The problem deals with finding roots of this type of equation. |
| Root of an Equation | A value of the variable (x) that makes the equation true (i.e., ${f(x) = 0}$). | We are looking for the interval containing such a value. |
| Function Evaluation | Calculating the value of ${f(x)}$ at a specific value of ${x}$. | Used to find ${f(0)}$ and ${f(2)}$ to check for sign changes. |
| Intermediate Value Theorem (IVT) | For a continuous function, if ${f(a)}$ and ${f(b)}$ have opposite signs, there's a root between ${a}$ and ${b}$. | The principle used to conclude a root exists in ${ (0, 2) }$. |
| Interval Notation | Representing a range of numbers on the number line, e.g., ${ (a, b) }$ for numbers between ${a}$ and ${b}$. | The solution is presented as an interval. |
Additional Information: Root Location and Function Values
Understanding the relationship between the function values at different points and the location of roots is a fundamental concept in algebra and calculus. For polynomial functions like quadratics, which are continuous everywhere, the sign of the function changes only at a root (or points where the graph touches the x-axis without crossing, but that's less common with sign changes). This is why evaluating ${f(x)}$ at specific points and checking the sign is a powerful tool for narrowing down where roots might be.
In this specific problem, the inequality ${4a + c < 2b}$ was key. By rearranging it, we found it directly gave us the sign of the function at ${x=2}$, i.g., ${f(2) < 0}$. Combined with the simple evaluation at ${x=0}$, which used the condition ${c > 0}$ to show ${f(0) > 0}$, the problem became straightforward.
This method of checking function signs at interval endpoints is very useful for locating roots, especially when an exact solution for the roots is difficult to find or not required.
Important Questions from Quadratic Equations
If k = c, then the roots of the equation are:
If \(\rm {k}=\frac{{c}}{2},({c} \neq 0)\), then the roots of the equation are :
What is the number of real roots of the equation?
What is the sum of all the roots of the equation?
If α and β are the distinct roots of equation x2 - x + 1 = 0, then what is the value of \(\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|\) ?
