Given that, the one root of $ax^{2}-bx+c = 0$ is $2+\sqrt{3}.$
We know that, if one root is $a+\sqrt{b},$ then another root will be $a-\sqrt{b}.$
And the quadratic equation, $ax^{2}+bx+c = 0$ has two roots $\alpha,$ and $\beta,$ then
- Sum of roots $= \alpha + \beta = \frac{-b}{a}$
- Product of roots $= \alpha \cdot \beta = \frac{c}{a}$
Let the quadratic equation, $ax^{2}-bx+c = 0$ have roots $\alpha = 2+ \sqrt{3}, \;\beta = 2- \sqrt{3}.$
So,
- $ \alpha + \beta = \frac{-(-b)}{a} = \frac{b}{a} = 2+\sqrt{3} + 2 – \sqrt{3} = 4 = \frac{4}{1}\quad \longrightarrow (1)$
- $\alpha \cdot \beta = \frac{c}{a} = (2+\sqrt{3}) (2 – \sqrt{3}) = 2^{2} – (\sqrt{3})^{2} = 1\quad \longrightarrow (2)$
Divide the equation $(1)$ by $(2)$
$\Rightarrow \dfrac{\frac{b}{a}}{\frac{c}{a}} = \frac{4}{1}$
$\Rightarrow \frac{b}{a} \times \frac{a}{c} = 4$
$\Rightarrow \frac{b}{c} = 4$
$\Rightarrow \frac{c^{3}}{c} = 4\quad [\because \text{Given that}, b= c^{3}] $
$\Rightarrow c^{2} = 4$
$\Rightarrow \boxed{c = 2}$
Put the value of $c$ in equation $(2),$ we get
$\Rightarrow \frac{c}{a} = 1$
$\Rightarrow \frac{2}{a} = 1$
$\Rightarrow \boxed{a = 2}$
$\Rightarrow \boxed{|a| = 2}$
Correct Answer $:\text{A}$
$\textbf{PS}:$ We know that,
$$|x| = \left\{\begin{matrix} x\;; &x\geq 0 \\ -x\;; & x< 0 \end{matrix}\right.$$
- $(a+b)(a-b) = a^{2} – b^{2}$