Given that $ax^2 + bx + c = 0$ is a quadratic equation with roots $5$ and $-1$, we can apply Vieta's formulas, which state:

1. The sum of the roots $r_1 + r_2 = -\frac{b}{a}$
2. The product of the roots $r_1 \cdot r_2 = \frac{c}{a}$

Step 1: Sum and Product of Roots

For the roots $r_1 = 5$ and $r_2 = -1$:

• The sum of the roots is $5 + (-1) = 4$
• The product of the roots is $5 \times (-1) = -5$

Step 2: Substitute into Standard Form

The general form of a quadratic equation is: $a(x - r_1)(x - r_2) = 0$

Substitute the roots $5$ and $-1$ into the factored form: $a(x - 5)(x + 1) = 0$

Step 3: Expand the Expression

Now, expand the factored form: $a(x - 5)(x + 1) = a(x^2 + x - 5x - 5) = a(x^2 - 4x - 5)$

Thus, the quadratic equation becomes: $a(x^2 - 4x - 5) = 0$

Step 4: Compare with Standard Form

By comparing the equation $a(x^2 - 4x - 5) = 0$ with the general form $ax^2 + bx + c = 0$, we can see:

• $b = -4a$
• $c = -5a$

Since $c$ is not given, $a$ can be any constant value. For simplicity, if we assume $a = 1$, then:

• $b = -4$
• $c = -5$

Thus, the values of $a$ and $b$ are: $a = 1, \quad b = -4$

Would you like more details or have any questions?

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Here are 5 related questions to deepen your understanding:

1. What happens if we change the sign of both roots?
2. How would the equation change if $a \neq 1$?
3. Can you derive the quadratic formula from Vieta’s formulas?
4. How would you solve this equation using the quadratic formula?
5. What are Vieta’s formulas used for in more advanced algebra?

Tip: Always check if factoring is the quickest method for solving quadratics before using the quadratic formula!