Consider the following for the next two (02) items that follow:
What is the value of α?
2
Solving Quadratic Equations: Finding the Value of Alpha
The question asks us to find the value of the constant $\alpha$ in the quadratic equation $2x^2 + 3x - \alpha = 0$, given that one of its roots is -2. We are also told that $\alpha > 0$ and the other root, $\beta$, is also positive ($\beta > 0$).
A fundamental property of equation roots is that they satisfy the equation. This means if $x=-2$ is a root of the equation $2x^2 + 3x - \alpha = 0$, then substituting $x=-2$ into the equation must make the equation true.
Step-by-Step Calculation for Alpha
- Start with the given quadratic equation: $2x^2 + 3x - \alpha = 0$.
- Substitute the known root, $x = -2$, into the equation. This will allow us to solve for the unknown $\alpha$.
Substituting $x = -2$ into the equation:
$$2(-2)^2 + 3(-2) - \alpha = 0$$
Now, we evaluate the terms:
$$2(4) + (-6) - \alpha = 0$$
$$8 - 6 - \alpha = 0$$
$$2 - \alpha = 0$$
To find $\alpha$, we isolate it:
$$\alpha = 2$$
Verifying the Conditions
The problem states that $\alpha > 0$. Our calculated value for $\alpha$ is 2, which is indeed greater than 0. This matches the condition given in the problem.
The information about the second equation ($x^2 – 3mx + 2m^2 = 0$) and the other root $\beta > 0$ from the first equation are additional details provided, possibly relevant for subsequent questions or to ensure the problem is well-defined. For finding the value of $\alpha$, knowing one root (-2) is sufficient.
Summary of Findings
By substituting the given root $x=-2$ into the equation $2x^2 + 3x - \alpha = 0$, we found that $\alpha = 2$. This value satisfies the condition $\alpha > 0$.
The value of $\alpha$ is 2.
| Equation | Given Root | Substitution | Result for $\alpha$ |
|---|---|---|---|
| $2x^2 + 3x - \alpha = 0$ | $x = -2$ | $2(-2)^2 + 3(-2) - \alpha = 0$ | $\alpha = 2$ |
Revision Table: Key Concepts for Quadratic Equations
| Concept | Description |
|---|---|
| Quadratic Equation | An equation of the form $ax^2 + bx + c = 0$, where $a \neq 0$. |
| Root of an Equation | A value of the variable that makes the equation true. If 'r' is a root of $P(x)=0$, then $P(r)=0$. |
| Sum of Roots | For $ax^2 + bx + c = 0$, the sum of roots is $-b/a$. |
| Product of Roots | For $ax^2 + bx + c = 0$, the product of roots is $c/a$. |
Additional Information: Properties of Quadratic Roots
For a quadratic equation $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$:
- Sum of roots: $r_1 + r_2 = -\frac{b}{a}$
- Product of roots: $r_1 \times r_2 = \frac{c}{a}$
In our given equation $2x^2 + 3x - \alpha = 0$, the roots are -2 and $\beta$. Comparing this to the standard form, we have $a=2$, $b=3$, and $c=-\alpha$.
Using the product of roots formula:
$$(-2) \times \beta = \frac{-\alpha}{2}$$
$$-2\beta = \frac{-\alpha}{2}$$
$$4\beta = \alpha$$
Since we found $\alpha = 2$, we can find $\beta$ using this relationship:
$$4\beta = 2$$
$$\beta = \frac{2}{4} = \frac{1}{2}$$
This confirms that $\beta = 1/2$, which is indeed greater than 0, consistent with the problem statement ($\beta > 0$). This method using root properties gives the same value for $\alpha$ as substituting the root, and also helps find the other root $\beta$.
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