if alpha and beta are the roots of the equation axsquare +bx+c=0 the find alpha/abeta+b  + beta/aalpha+b  Solve through quadratic equation

Given the quadratic equation \( ax^2 + bx + c = 0 \), where \(\alpha\) and \(\beta\) are the roots, we are required to find the value of the following expression:

\[
\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b}
\]

### Step 1: Use Vieta's formulas
From Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \), the sum and product of the roots are given by:

\[
\alpha + \beta = -\frac{b}{a}
\]
\[
\alpha \beta = \frac{c}{a}
\]

### Step 2: Simplify the given expression
We start by simplifying the expression:

\[
\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b}
\]

To simplify this, we need a common denominator:

\[
\text{Common denominator} = (a\beta + b)(a\alpha + b)
\]

Thus, the expression becomes:

\[
\frac{\alpha(a\alpha + b) + \beta(a\beta + b)}{(a\beta + b)(a\alpha + b)}
\]

Expanding the numerator:

\[
\alpha(a\alpha + b) + \beta(a\beta + b) = a\alpha^2 + b\alpha + a\beta^2 + b\beta
\]

We can group terms:

\[
a(\alpha^2 + \beta^2) + b(\alpha + \beta)
\]

### Step 3: Simplify \(\alpha^2 + \beta^2\)
The expression \(\alpha^2 + \beta^2\) can be written in terms of \(\alpha + \beta\) and \(\alpha \beta\):

\[
\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta
\]

Substitute the values from Vieta's formulas:

\[
\alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2 \times \frac{c}{a} = \frac{b^2}{a^2} - \frac{2c}{a}
\]

So the numerator becomes:

\[
a\left(\frac{b^2}{a^2} - \frac{2c}{a}\right) + b\left(-\frac{b}{a}\right) = \frac{ab^2}{a^2} - 2ac + \frac{-b^2}{a}
\]

Simplifying:

\[
\frac{b^2}{a} - 2ac - \frac{b^2}{a} = -2ac
\]

### Step 4: Denominator Simplification
The denominator is:

\[
(a\beta + b)(a\alpha + b) = a^2\alpha \beta + ab(\alpha + \beta) + b^2
\]

Substitute \(\alpha \beta = \frac{c}{a}\) and \(\alpha + \beta = -\frac{b}{a}\):

\[
a^2\frac{c}{a} + ab\left(-\frac{b}{a}\right) + b^2 = ac - b^2 + b^2 = ac
\]

### Step 5: Final expression
Finally, the expression simplifies to:

\[
\frac{-2ac}{ac} = -2
\]

So, the value of the expression is \(\boxed{-2}\).

Would you like more details on any step, or have any questions?

Here are some related questions:
1. How does Vieta's formula relate to the roots of quadratic equations?
2. What are the conditions for the discriminant of a quadratic equation?
3. How do you derive the formula for \(\alpha^2 + \beta^2\) using \(\alpha + \beta\) and \(\alpha \beta\)?
4. Can you generalize the result for different forms of quadratic equations?
5. What is the geometric interpretation of the roots of a quadratic equation?

**Tip:** Always check if you can simplify expressions involving roots by using Vieta's formulas, which relate the coefficients of the polynomial to the sum and product of its roots.

Learn how to simplify and solve the expression (alpha/a*beta+b) + (beta/a*alpha+b) using Vieta's formulas and algebraic simplification. This problem involves quadratic equations and Vieta's formulas, suitable for advanced high school students.

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