Suppose one of the roots of the equation $ax^2 - bx + c = 0$ is $2+\sqrt{3}$ where a, b and c are rational numbers and $a \neq 0$. If $b = c^3$ then $|a|$ equals

When a quadratic equation has rational coefficients (like integers or fractions) and one root contains a square root, the other root must be its conjugate.

Since one root is $2 + \sqrt{3}$, the other root must be $2 - \sqrt{3}$.

When we have rational coefficients, irrational roots always appear in conjugate pairs. This ensures that when we multiply out the factors, all the "messy" square root terms cancel out, leaving us with rational coefficients.

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For any quadratic $ax^2 - bx + c = 0$, the sum of roots equals $\dfrac{b}{a}$.

Sum of roots = $(2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$

Notice how the $\sqrt{3}$ terms cancel out perfectly!

Therefore: $\dfrac{b}{a} = 4$ ... (1)

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For any quadratic $ax^2 - bx + c = 0$, the product of roots equals $\dfrac{c}{a}$.

Product of roots = $(2 + \sqrt{3})(2 - \sqrt{3})$

Using the difference of squares formula: $(a + b)(a - b) = a^2 - b^2$

$(2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$

Therefore: $\dfrac{c}{a} = 1$ ... (2)

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Dividing equation (1) by equation (2):

$\dfrac{b/a}{c/a} = \dfrac{4}{1}$

$\dfrac{b}{c} = 4$

Therefore: $b = 4c$ ... (3)

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We're told that $b = c^3$.

Combining with equation (3): $c^3 = 4c$

$c^3 - 4c = 0$

$c(c^2 - 4) = 0$

$c(c - 2)(c + 2) = 0$

So $c = 0$, $c = 2$, or $c = -2$.

If $c = 0$, then from equation (2), $\dfrac{c}{a} = 1$ would mean $0 = 1$, which is impossible.

Therefore: $c = 2$ or $c = -2$

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Case 1: $c = 2$

$b = c^3 = 2^3 = 8$

From $\dfrac{b}{a} = 4$: $\dfrac{8}{a} = 4$, so $a = 2$

Therefore $|a| = 2$

Case 2: $c = -2$

$b = c^3 = (-2)^3 = -8$

From $\dfrac{b}{a} = 4$: $\dfrac{-8}{a} = 4$, so $a = -2$

Therefore $|a| = 2$

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In both valid cases, $|a| = 2$.

When dealing with quadratic equations with rational coefficients and irrational roots, we should always remember that the roots come in conjugate pairs. This makes the sum and product calculations much cleaner!