Imagine imaginary!

Algebra Level 2

Consider a quadratic equation q x 2 + 2 p x + 2 q = 0 qx^2 +2px +2q =0 having real and unequal roots. Is it true that roots of the equation ( p + q ) x 2 + 2 q x + ( p − q ) = 0 \,(p+q)x^2 +2qx +(p-q) =0 are imaginary ?

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Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Tom Engelsman
Mar 18, 2021

The roots of the first quadratic equation equal:

x = − 2 p ± 4 p 2 − 4 ( q ) ( 2 q ) 2 q = − 2 p ± 4 ( p 2 − 2 q 2 ) 2 q = − p ± p 2 − 2 q 2 q x = \frac{-2p \pm \sqrt{4p^2 - 4(q)(2q)}}{2q} = \frac{-2p \pm \sqrt{4(p^2-2q^2)}}{2q} = \frac{-p \pm \sqrt{p^2-2q^2}}{q}

and if the roots are real and distinct ⇒ p 2 − 2 q 2 > 0. \Rightarrow p^2 - 2q^2 > 0. Now, the roots of the second quadratic equation equal:

x = − 2 q ± 4 q 2 − 4 ( p + q ) ( p − q ) 2 ( p + q ) = − q ± q 2 − p 2 + q 2 p + q = − q ± − ( p 2 − 2 q 2 ) p + q x = \frac{-2q \pm \sqrt{4q^2 - 4(p+q)(p-q)}}{2(p+q)} = \frac{-q \pm \sqrt{q^2 - p^2 + q^2}}{p+q} = \frac{-q \pm \sqrt{-(p^2-2q^2)}}{p+q}

and since we know that p 2 − 2 q 2 > 0 ⇒ p^2 - 2q^2 > 0 \Rightarrow the roots of this second quadratic are imaginary due to a negative-valued discriminant.

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