Question

If the equation \(a^{2} t^{2}-a c t+a=0\) has roots \(\alpha\) and \(\beta\), then the value of \((\alpha-\beta)^{2}\) is (a) \(\frac{3 \mathrm{ac}+\mathrm{a}^{2}}{2 \mathrm{c}}\) (b) \(\frac{2 \mathrm{a}+\mathrm{c}}{\mathrm{a}^{2}}\) (c) \(\frac{4 a^{2}+c^{2}}{2 a}\) (d) \(\frac{c^{2}-4 a}{a^{2}}\)

Short answer

Answer: \((\alpha-\beta)^{2} = \frac{c^{2}-4a}{a^{2}}\)

Step-by-step solution

Find the sum of the roots using Vieta's formulas

Vieta's formulas relate the coefficients of a quadratic equation to the sum and product of its roots. For a quadratic equation of the form \(at^{2} + bt + c = 0\), the sum of the roots (\(α + β\)) is given by the formula: $\displaystyle \alpha+\beta=-\frac{b}{a} $ Here, our original equation is \(a^{2}t^{2}-act+a=0\). So, \(a = a^2\), \(b = -ac\), and \(c = a\). Now let's calculate the sum of the roots: $\displaystyle \alpha+\beta=-\frac{-ac}{a^{2}}=\frac{ac}{a^{2}} $

Find the product of the roots using Vieta's formulas

Similarly, for a quadratic equation of the form \(at^{2} + bt + c = 0\), the product of the roots (\(αβ\)) is given by the formula: $\displaystyle \alpha\beta=\frac{c}{a} $ Using \(a = a^2\), \(b = -ac\), and \(c = a\), let's calculate the product of the roots: $\displaystyle \alpha\beta=\frac{a}{a^{2}}=\frac{1}{a} $

Use the sum and product to find \((α-β)^{2}\)

We know that \((\alpha-\beta)^{2}=(\alpha+\beta)^{2} - 4\alpha\beta\). Substitute the values for \(\alpha+\beta\) and \(\alpha\beta\) found in Step 1 and Step 2: $ (\alpha-\beta)^{2}=\left(\frac{ac}{a^{2}}\right)^{2}-4\left(\frac{1}{a}\right) $

Simplify the expression

Now let's simplify the expression to get a final value for \((\alpha-\beta)^{2}\): $ (\alpha-\beta)^{2}=\frac{a^2c^2}{a^{4}}-\frac{4}{a} $ In order to have a common denominator, multiply the second term by \(\frac{a^3}{a^3}\): $ (\alpha-\beta)^{2}=\frac{a^2c^2}{a^{4}}-\frac{4a^3}{a^4} $ Now combine the terms: $ (\alpha-\beta)^{2}=\frac{a^2c^2-4a^3}{a^4} $ Looking at the given options, it matches with option (d). Therefore, the value of \((\alpha-\beta)^{2}\) is: $ (\alpha-\beta)^{2}=\boxed{\frac{c^{2}-4a}{a^{2}}} $

Key concepts

Vieta's Formulas

Vieta's formulas are incredibly useful in solving quadratic equations. They provide a direct relationship between the coefficients of the quadratic equation and the roots. Let's break down how this works.

Consider a quadratic equation of the form \(at^{2} + bt + c = 0\). According to Vieta's formulas:

• The sum of the roots \(\alpha + \beta\) is \(-\frac{b}{a}\).
• The product of the roots \(\alpha \beta\) is \(\frac{c}{a}\).

In this problem, we have the equation \(a^{2}t^{2}-act+a=0\), where the coefficients are \(a = a^2\), \(b = -ac\), and \(c = a\). By applying Vieta's formulas:

• The sum of the roots \(\alpha + \beta = \frac{ac}{a^{2}}\).
• The product of the roots \(\alpha \beta = \frac{1}{a}\).

These relationships simplify working with the roots and are crucial for further calculations in quadratic problems.

Roots of Equations

The roots of a quadratic equation are the values of \(t\) that satisfy the equation \(at^{2} + bt + c = 0\). In other words, they're the solutions to the equation. Here's what you need to know about roots:

• Each quadratic equation has exactly two roots, which can be real or complex numbers.
• The roots can be found using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• In some cases, Vieta's formulas provide a faster way to understand properties of the roots without solving the quadratic formula.

For the given quadratic equation \(a^{2}t^{2}-act+a=0\), you can use Vieta's formulas to quickly identify the sum and product of its roots, as shown earlier. This understanding is key for comparing how roots relate to each other, or deriving further expressions involving roots.

Sum and Product of Roots

Understanding the sum and product of roots isn't just about memorizing formulas—it's about recognizing their practical applications in solving problems. When you have a quadratic equation, these concepts simplify comparisons and derivations.

Using the quadratic equation \(at^{2} + bt + c = 0\), remember:

• Sum of the roots: \(\alpha + \beta = -\frac{b}{a}\).
• Product of the roots: \(\alpha \beta = \frac{c}{a}\).

For the specific equation \(a^{2}t^{2}-act+a=0\), you calculate:

• The sum of \(\alpha + \beta = \frac{ac}{a^{2}}\).
• The product \(\alpha \beta = \frac{1}{a}\).

These results are particularly useful when you want to find expressions like \((\alpha - \beta)^2\). For example, using the identity \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta\), you substitute the calculated values and simplify to find the final expression for \((\alpha - \beta)^2\). This approach streamlines the process and allows you to solve the problem efficiently.