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Question

Mathematics Question on Complex Numbers and Quadratic Equations

Let p,q,rRp, q, r \,\in\, R and r>p>0r > p > 0. If the quadratic equation px2+qx+r=0px^2 + qx + r = 0 has two complex roots α\alpha and β\beta, then α+β\left|\alpha\right|+\left|\beta\right| is

A

equal to 1

B

less than 2 but not equal to 1

C

greater than 2

D

equal to 2

Answer

greater than 2

Explanation

Solution

Given quadratic equation is px2+qcx+r=0...(1)px^{2}+qcx+r=0\,...\left(1\right) D=q24prD=q^{2}-4pr Since α\alpha and β\beta are two complex root β=αˉβ=αˉβ=α\therefore \beta=\bar{\alpha} \Rightarrow \left|\beta\right|=\left|\bar{\alpha}\right| \Rightarrow \left|\beta\right|=\left|\alpha\right| (αˉ=α)\left(\because \left|\bar{\alpha}\right|=\left|\alpha\right|\right) Consider α+β=α+α(β=α)\left|\alpha\right|+\left|\beta\right|=\left|\alpha\right|+\left|\alpha\right|\, \left(\because\left|\beta\right|=\left|\alpha\right|\right) =2α>2.1=2(αˉ>1)=2\left|\alpha\right| > 2.1=2\,\left(\because \left|\bar{\alpha }\right|> 1\right) Hence, α+β\left|\alpha \right|+\left|\beta \right| is greater than 2.2.