# CAT 2021 Set-2 | Quantitative Aptitude | Question: 16

1 vote
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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

1. $2$
2. $4$
3. $1$
4. $3$

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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,

$$|x| = \left\{\begin{matrix} x\;; &x\geq 0 \\ -x\;; & x< 0 \end{matrix}\right.$$

•  $(a+b)(a-b) = a^{2} – b^{2}$
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