The quadratic equation $x² + bx + c = 0$ has two roots $4a$ and $3a,$ where a is an integer. Which of the following is a possible value of $b² + c$?

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We have a quadratic equation $x^2 + bx + c = 0$ with two specific roots: $4a$ and $3a$, where $a$ is an integer.

We need to find a possible value of $b^2 + c$.

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For any quadratic equation $x^2 + bx + c = 0$ with roots $r_1$ and $r_2$, Vieta's formulas tell us:

Sum of roots: $r_1 + r_2 = -b$

Product of roots: $r_1 \times r_2 = c$

When we factor a quadratic with roots $r_1$ and $r_2$, we get $(x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1r_2$. Comparing with $x^2 + bx + c$, we see that $b = -(r_1 + r_2)$ and $c = r_1r_2$.

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Our roots are $4a$ and $3a$, so:

Sum of roots: $4a + 3a = 7a = -b$

Therefore: $b = -7a$

Product of roots: $(4a) \times (3a) = 12a^2 = c$

Therefore: $c = 12a^2$

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Now we can substitute:

$b^2 + c = (-7a)^2 + 12a^2$

$b^2 + c = 49a^2 + 12a^2$

$b^2 + c = 61a^2$

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Since $a$ is an integer, $a$ can be any whole number: ..., -2, -1, 0, 1, 2, ...

This means $a^2$ can be: 0, 1, 4, 9, 16, 25, ...

Therefore, possible values of $b^2 + c = 61a^2$ are:

When $a = 0$: $b^2 + c = 61(0) = 0$

When $a = ±1$: $b^2 + c = 61(1) = 61$

When $a = ±2$: $b^2 + c = 61(4) = 244$

When $a = ±3$: $b^2 + c = 61(9) = 549$

And so on...

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Any value of the form $61k$ where $k$ is a perfect square (0, 1, 4, 9, 16, ...) is a possible value of $b^2 + c$.

The most common possible values are: 0, 61, 244, 549, 976, ...