Understanding the Problem: Quadratic Equations and Shared Roots

We are given a quadratic equation $x^2 + 3x + 2 = 0$. We are told that both roots of this equation are also roots of another quadratic equation, $ax^2 + bx + c = 0$. We need to determine what must be true about the second equation.

Finding the Roots of the First Quadratic Equation

First, let's find the roots of the given quadratic equation $x^2 + 3x + 2 = 0$. We can solve this by factoring the quadratic expression:

We look for two numbers that multiply to 2 and add up to 3. These numbers are 1 and 2.

So, we can factor the equation as:

$$ (x+1)(x+2) = 0 $$

This equation holds true if $x+1 = 0$ or $x+2 = 0$. Therefore, the roots are:

• $x = -1$
• $x = -2$

Implications of Shared Roots

The problem states that these roots, $-1$ and $-2$, are also the roots of the second quadratic equation $ax^2 + bx + c = 0$.

If a quadratic equation has roots $r_1$ and $r_2$, it can be written in the form $k(x - r_1)(x - r_2) = 0$, where $k$ is a non-zero constant.

Since the roots are $-1$ and $-2$, the second quadratic equation $ax^2 + bx + c = 0$ must be equivalent to:

$$ k(x - (-1))(x - (-2)) = 0 $$ $$ k(x+1)(x+2) = 0 $$

Expanding the factored form:

$$ k(x^2 + 2x + x + 2) = 0 $$ $$ k(x^2 + 3x + 2) = 0 $$

So, the second quadratic equation must be of the form:

$$ kx^2 + 3kx + 2k = 0 $$

Comparing this with the given equation $ax^2 + bx + c = 0$, we can see the relationship between the coefficients:

• $a = k$
• $b = 3k$
• $c = 2k$

Note that for $ax^2 + bx + c = 0$ to be a quadratic equation, $a$ must be non-zero, which means $k$ must also be non-zero.

Analyzing the Options

Let's examine each option based on the relationship $a=k$, $b=3k$, and $c=2k$:

1. Option 1: $a+3b+2c = 0$

   Substitute the values: $k + 3(3k) + 2(2k) = k + 9k + 4k = 14k$. Since $k$ is non-zero, $14k$ is not necessarily 0. So, this option is not always true.

2. Option 2: $a=b=c$

   Substitute the values: $k = 3k = 2k$. This equality only holds if $k=0$. However, $a$ cannot be zero for a quadratic equation. So, this option is not true.

3. Option 3: $\frac{a}{1} = \frac{b}{2} = \frac{c}{3}$

   Substitute the values: $\frac{k}{1} = \frac{3k}{2} = \frac{2k}{3}$. This simplifies to $k = \frac{3}{2}k$ and $k = \frac{2}{3}k$. Both conditions imply $k=0$, which is not allowed. So, this option is not true.

4. Option 4: $ax^2 + bx + c$ must be a multiple of $x^2 + 3x+2$

   We found that $ax^2 + bx + c = kx^2 + 3kx + 2k = k(x^2 + 3x + 2)$. This shows that the polynomial $ax^2 + bx + c$ is exactly $k$ times the polynomial $x^2 + 3x + 2$. Therefore, $ax^2 + bx + c$ is a multiple of $x^2 + 3x + 2$, where the multiplier is $k$. This statement must be true.

Conclusion

Based on the analysis, the only statement that must be true is that the second quadratic polynomial $ax^2 + bx + c$ is a multiple of the first quadratic polynomial $x^2 + 3x + 2$.