IIT JEE 1982 Maths - 'Adapted' - FIB to Single-Digit Integer Q6

Algebra Level 3

If $2+i\sqrt3$ is root of the equation $x^2+px+q=0$ , where $p$ and $q$ are real, then find $\left \lfloor \dfrac {\lvert p\rvert+\lvert q\rvert}2 \right \rfloor$ .

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Notations:

$i = \sqrt {-1}$ denotes the imaginary unit .
$\lfloor \cdot \rfloor$ denotes the floor function .
$| \cdot |$ denotes the absolute value function .

If you are preparing for IIT JEE, you can try IIT JEE 1982 Mathematics Archives .

The answer is 5.

1 solution

Chew-Seong Cheong
Feb 2, 2017

If $\alpha = 2 + i\sqrt 3$ is a root, we can expext $\overline \alpha = 2 - i\sqrt 3$ , its conjugate is also a root so that the quadratic equation has real coefficients. By Vieta's formula , we have:

$\begin{cases} \alpha + \overline \alpha = - p & \implies 2 + i\sqrt 3 + 2 - i\sqrt 3 = 4 & \implies p = - 4 \\ \alpha \overline \alpha = q & \implies (2 + i\sqrt 3)(2 - i\sqrt 3) = 7 & \implies q = 7 \end{cases}$

$\implies \left \lfloor \dfrac {|p|+|q|}2 \right \rfloor = \left \lfloor \dfrac {4+7}2 \right \rfloor = \boxed{5}$

Did the same way

I Gede Arya Raditya Parameswara - 4 years, 4 months ago

IIT JEE 1982 Maths - 'Adapted' - FIB to Single-Digit Integer Q6

If you are preparing for IIT JEE, you can try IIT JEE 1982 Mathematics Archives .

The answer is 5.

1 solution

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