Question:
Let $p, q, r \,\in\, R$ and $r > p > 0$. If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta$, then $\left|\alpha\right|+\left|\beta\right|$ is
Let $p, q, r \,\in\, R$ and $r > p > 0$. If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta$, then $\left|\alpha\right|+\left|\beta\right|$ is
Updated On: Jul 28, 2022
- equal to 1
- less than 2 but not equal to 1
- greater than 2
- equal to 2
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The Correct Option is C
Solution and Explanation
Given quadratic equation is
$px^{2}+qcx+r=0\,...\left(1\right)$
$D=q^{2}-4pr$
Since $\alpha$ and $\beta$ are two complex root
$\therefore \beta=\bar{\alpha} \Rightarrow \left|\beta\right|=\left|\bar{\alpha}\right| \Rightarrow \left|\beta\right|=\left|\alpha\right|$
$\left(\because \left|\bar{\alpha}\right|=\left|\alpha\right|\right)$
Consider
$\left|\alpha\right|+\left|\beta\right|=\left|\alpha\right|+\left|\alpha\right|\, \left(\because\left|\beta\right|=\left|\alpha\right|\right)$
$=2\left|\alpha\right| > 2.1=2\,\left(\because \left|\bar{\alpha }\right|> 1\right)$
Hence, $\left|\alpha \right|+\left|\beta \right|$ is greater than $2.$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
