Solving For Beta Squared Roots Of Quadratic Equation

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This mathematical problem delves into the fascinating realm of quadratic equations and their roots. Specifically, we are presented with a quadratic equation, $a^2 x^2 + x + 1 - a^2 = 0$, and told that its roots are $\alpha^2$ and $-\beta^2$. Our mission is to determine the value of $\beta^2$. To embark on this mathematical journey, we'll leverage the fundamental relationships between the roots and coefficients of a quadratic equation.

Understanding the Roots and Coefficients Relationship

At the heart of our problem lies the connection between the roots and coefficients of a quadratic equation. Consider a general quadratic equation of the form $Ax^2 + Bx + C = 0$. If the roots of this equation are $r_1$ and $r_2$, then the following relationships hold:

• Sum of roots: $r_1 + r_2 = -B/A$
• Product of roots: $r_1 * r_2 = C/A$

These relationships are derived from Vieta's formulas and provide a powerful tool for analyzing quadratic equations. In essence, they tell us that the sum and product of the roots are directly related to the coefficients of the quadratic equation. This connection is crucial for solving our problem.

Applying the Relationships to Our Problem

Now, let's apply these relationships to the given quadratic equation, $a^2 x^2 + x + 1 - a^2 = 0$. We are given that the roots are $\alpha^2$ and $-\beta^2$. Therefore, we can write:

• $r_1 = \alpha^2$
• $r_2 = -\beta^2$

Identifying the coefficients in our equation, we have:

• $A = a^2$
• $B = 1$
• $C = 1 - a^2$

Using the sum of roots relationship, we get:

$\alpha^2 + (-\beta^2) = -1/a^2$
$\alpha^2 - \beta^2 = -1/a^2$ (Equation 1)

Next, using the product of roots relationship, we get:

$\alpha^2 * (-\beta^2) = (1 - a^2) / a^2$
$-\alpha^2 \beta^2 = (1 - a^2) / a^2$ (Equation 2)

Manipulating the Equations

We now have two equations with two unknowns, $\alpha^2$ and $\beta^2$. Our goal is to solve for $\beta^2$. To achieve this, we need to manipulate these equations strategically. Let's start by isolating $\alpha^2$ in Equation 1:

$\alpha^2 = \beta^2 - 1/a^2$ (Equation 3)

Now, we can substitute this expression for $\alpha^2$ into Equation 2:

$-(\beta^2 - 1/a^2) \beta^2 = (1 - a^2) / a^2$

Expanding and rearranging the terms, we get:

$-\beta^4 + \beta^2/a^2 = (1 - a^2) / a^2$

Multiplying both sides by $-1$: $\beta^4 - \beta^2/a^2 = (a^2 - 1) / a^2$
Now, let's multiply both sides of the equation by $a^2$ to eliminate the fractions:

$a^2 \beta^4 - \beta^2 = a^2 - 1$

Rearranging the terms, we obtain a quadratic equation in $\beta^2$:

$a^2 \beta^4 - \beta^2 - (a^2 - 1) = 0$

Let $y = \beta^2$, the equation becomes:

$a^2 y^2 - y - (a^2 - 1) = 0$

This is a quadratic equation in terms of $y = \beta^2$. We can solve for $y$ using the quadratic formula, factoring, or other suitable methods.

Solving for $\beta^2$

To solve the quadratic equation $a^2 y^2 - y - (a^2 - 1) = 0$, we can use factoring technique. Let's rewrite the quadratic equation as:

$a^2 y^2 - a^2y + a^2y - y - (a^2 - 1) = 0$
We can factor this quadratic equation by splitting the middle term. We are looking for two numbers that multiply to $a^2(-(a^2-1)) = -a^4+a^2$ and add up to $-1$. We can observe that the equation can be factored directly:

$a^2 y^2 - y - (a^2 - 1) = 0$

We can use the quadratic formula to find the roots of this equation. Recall that for a quadratic equation of the form $Ax^2 + Bx + C = 0$, the solutions are given by:

$x = [-B ± sqrt(B^2 - 4AC)] / (2A)$

In our case, we have $A = a^2$, $B = -1$, and $C = -(a^2 - 1)$. Plugging these values into the quadratic formula, we get:

$y = [1 ± sqrt((-1)^2 - 4 * a^2 * -(a^2 - 1))] / (2a^2)$
$y = [1 ± sqrt(1 + 4a^4 - 4a^2)] / (2a^2)$

Notice that the term inside the square root can be rewritten as a perfect square:

$1 + 4a^4 - 4a^2 = (2a^2 - 1)^2$

Therefore, we have:

$y = [1 ± sqrt((2a^2 - 1)^2)] / (2a^2)$
$y = [1 ± (2a^2 - 1)] / (2a^2)$

This gives us two possible solutions for $y$:

$y_1 = [1 + (2a^2 - 1)] / (2a^2) = 2a^2 / (2a^2) = 1$
$y_2 = [1 - (2a^2 - 1)] / (2a^2) = (2 - 2a^2) / (2a^2) = (1 - a^2) / a^2$

Since $y = \beta^2$, we have two possible values for $\beta^2$:

$\beta^2 = 1$ or $\beta^2 = (1 - a^2) / a^2$

Let's substitute $\beta^2 = 1$ into Equation 3:

$\alpha^2 = 1 - 1/a^2 = (a^2 - 1) / a^2$
Substitute $\beta^2 = 1$ and $\alpha^2 = (a^2 - 1) / a^2$ into Equation 2:

$-((a^2 - 1) / a^2) * 1 = (1 - a^2) / a^2$
$-(a^2 - 1) / a^2 = (1 - a^2) / a^2$
$(1 - a^2) / a^2 = (1 - a^2) / a^2$

This equality holds, so $\beta^2 = 1$ is a valid solution.

Now let's consider the option $\beta^2 = (1 - a^2) / a^2$. Substituting this into Equation 3: $\alpha^2 = (1 - a^2) / a^2 - 1/a^2 = (1 - a^2 - 1) / a^2 = -1$
This is a valid option if $\alpha^2$ can be $-1$.

Substitute $\beta^2 = (1 - a^2) / a^2$ and $\alpha^2 = -1$ into Equation 2: $-(-1) ((1 - a^2) / a^2) = (1 - a^2) / a^2$
$(1 - a^2) / a^2 = (1 - a^2) / a^2$

This equality also holds. However, if we look at the answer choices given, we have the option $\beta^2 = 1$. Therefore, let's analyze if that fits.

The Solution: $\beta^2 = 1$

Therefore, based on our analysis and the provided options, the solution is:

$\beta^2 = 1$

This problem highlights the power of Vieta's formulas and the relationships between the roots and coefficients of a quadratic equation. By carefully applying these relationships and manipulating the resulting equations, we were able to successfully determine the value of $\beta^2$.

Final Answer: The final answer is $\boxed{1}$