If α and β are the roots of the equation ax 2+ bx + c = 0, then the value of 1/(aα + b) + 1/(aβ + b) is
b/ac
Understanding the Quadratic Equation and its Roots
The question asks us to find the value of a specific expression involving the roots of a quadratic equation. A quadratic equation is an equation of the form \(\(ax^2 + bx + c = 0\)\), where \(a\), \(b\), and \(c\) are coefficients, and \(a \neq 0\). The roots of this equation, typically denoted by \(\(\alpha\)\) and \(\(\beta\)\), are the values of \(x\) that satisfy the equation.
Properties of Roots of a Quadratic Equation
For a quadratic equation \(\(ax^2 + bx + c = 0\)\), the sum and product of its roots \(\(\alpha\)\) and \(\(\beta\)\) are given by Vieta's formulas:
- Sum of roots: \(\(\alpha + \beta = -\frac{b}{a}\)\)
- Product of roots: \(\(\alpha \beta = \frac{c}{a}\)\)
Evaluating the Expression 1/(aα + b) + 1/(aβ + b)
We are given that \(\(\alpha\)\) and \(\(\beta\)\) are roots of \(\(ax^2 + bx + c = 0\)\). This means they satisfy the equation:
- For root \(\(\alpha\)\): \(\(a\alpha^2 + b\alpha + c = 0\)\)
- For root \(\(\beta\)\): \(\(a\beta^2 + b\beta + c = 0\)\)
We need to find the value of \(\(\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}\)\). Let's manipulate the equations from the roots to find expressions for \(\(a\alpha + b\)\) and \(\(a\beta + b\)\).
Consider the equation for root \(\(\alpha\)\): \(\(a\alpha^2 + b\alpha + c = 0\)\).
We can factor out \(\(\alpha\)\) from the first two terms:
\(\alpha(a\alpha + b) + c = 0\)
\(\alpha(a\alpha + b) = -c\)
Assuming \(\(\alpha \neq 0\)\) (if \(\(\alpha = 0\)\), then \(\(c=0\)\), and the equation would be \(\(ax^2+bx=0\)\), roots \(0\) and \(-b/a\). If \(c=0\), then \(b/ac\) is undefined. Let's assume \(c \neq 0\), which implies roots are non-zero unless \(b=0\) as well, in which case \(ax^2=0\) with root 0, but then \(c\) must be 0 for 0 to be a root). So, we can write:
\(a\alpha + b = -\frac{c}{\alpha}\)
Similarly, for the root \(\(\beta\)\) from \(\(a\beta^2 + b\beta + c = 0\)\):
\(\beta(a\beta + b) + c = 0\)
\(\beta(a\beta + b) = -c\)
\(a\beta + b = -\frac{c}{\beta}\)
Now substitute these expressions into the sum we need to evaluate:
\(\frac{1}{a\alpha + b} + \frac{1}{a\beta + b} = \frac{1}{-c/\alpha} + \frac{1}{-c/\beta}\)
\(= -\frac{\alpha}{c} - \frac{\beta}{c}\)
\(= -\left(\frac{\alpha + \beta}{c}\right)\)
We know from Vieta's formulas that the sum of the roots \(\(\alpha + \beta = -\frac{b}{a}\)\). Substitute this into the expression:
\(= -\left(\frac{-b/a}{c}\right)\)
\(= -\left(-\frac{b}{ac}\right)\)
\(= \frac{b}{ac}\)
Thus, the value of the expression \(\(\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}\)\) is \(\(\frac{b}{ac}\)\).
Step-by-Step Derivation Summary
- Start with the quadratic equation \(\(ax^2 + bx + c = 0\)\) and its roots \(\(\alpha\), \(\beta\)\).
- Use the property that roots satisfy the equation: \(\(a\alpha^2 + b\alpha + c = 0\)\) and \(\(a\beta^2 + b\beta + c = 0\)\).
- Rearrange these equations to find expressions for \(\(a\alpha + b\)\) and \(\(a\beta + b\)\) in terms of \(\(\alpha\), \(\beta\), c\): \(\(a\alpha + b = -c/\alpha\)\) and \(\(a\beta + b = -c/\beta\)\).
- Substitute these expressions into the target expression: \(\(\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}\)\).
- Simplify the resulting expression: \(\(-\frac{\alpha}{c} - \frac{\beta}{c} = -\frac{\alpha + \beta}{c}\)\).
- Substitute the sum of roots formula \(\(\alpha + \beta = -b/a\)\).
- Calculate the final value: \(\(-\frac{-b/a}{c} = \frac{b}{ac}\)\).
| Step | Process | Result |
|---|---|---|
| 1 | Start with root equations | \(\(a\alpha^2 + b\alpha + c = 0\)\), \(\(a\beta^2 + b\beta + c = 0\)\) |
| 2 | Rearrange for \(\(a\alpha+b\), \(a\beta+b\)\) | \(\(a\alpha + b = -c/\alpha\)\), \(\(a\beta + b = -c/\beta\)\) |
| 3 | Substitute into expression | \(\(\frac{1}{-c/\alpha} + \frac{1}{-c/\beta}\)\) |
| 4 | Simplify the sum | \(\(-\frac{\alpha}{c} - \frac{\beta}{c} = -\frac{\alpha + \beta}{c}\)\) |
| 5 | Use sum of roots formula | \(\(\alpha + \beta = -b/a\)\) |
| 6 | Final Calculation | \(\(-\frac{-b/a}{c} = \frac{b}{ac}\)\) |
Revision Table: Key Concepts for Quadratic Equations
| Concept | Description | Formula (for \(\(ax^2+bx+c=0\)\)) |
|---|---|---|
| Quadratic Equation | A polynomial equation of the second degree. | \(\(ax^2+bx+c=0\)\) (\(\(a \neq 0\)\)) |
| Roots of Equation | The values of the variable that satisfy the equation. | \(\(\alpha, \beta\)\) |
| Sum of Roots | The sum of the two roots. | \(\(\alpha + \beta = -b/a\)\) |
| Product of Roots | The product of the two roots. | \(\(\alpha \beta = c/a\)\) |
| Discriminant | Determines the nature of the roots. | \(\(\Delta = b^2 - 4ac\)\) |
Additional Information: Extending Root Properties
Understanding the relationship between the coefficients and roots of a polynomial equation is fundamental in algebra. Vieta's formulas provide a powerful tool for solving problems involving symmetric expressions of roots without actually finding the values of the roots themselves. This approach is particularly useful in competitive exams.
For instance, if we were asked to find \(\(\alpha^2 + \beta^2\)\), we could express it in terms of the sum and product of roots:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
Substituting the formulas for sum and product:
\(\alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right)\)
\(= \frac{b^2}{a^2} - \frac{2c}{a}\)
\(= \frac{b^2 - 2ac}{a^2}\)
This demonstrates how various expressions involving roots can be simplified using Vieta's formulas, similar to how we solved the original problem.
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