Complex Quadratics

Algebra Level pending

The quadratic equation p ( x ) = 0 p(x)=0 having real coefficients has purely imaginary roots. Then the equation p ( p ( x ) ) = 0 p(p(x))=0 has :-

Two real and two purely imaginary roots Only purely imaginary roots Neither real nor purely imaginary roots All real roots

Parth Sankhe by Parth Sankhe , 27, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Jordan Cahn
Dec 7, 2018

As an example, consider the equation p ( x ) = x 2 + 1 p(x) = x^2 + 1 , which has purely imaginary roots x = ± i x=\pm i . Then p ( p ( x ) ) = ( x 2 + 1 ) 2 + 1 = x 4 + 2 x 2 + 2 p(p(x)) = \left(x^2 + 1\right)^2 + 1 = x^4 + 2x^2 + 2 This is a biquadratic. By the quadratic formula, x 2 = − 2 ± 4 − 8 2 = − 1 ± i x^2 = \frac{-2\pm\sqrt{4-8}}{2} = -1 \pm i

Thus x = ± − 1 + i x=\pm\sqrt{-1 + i} or x = ± − 1 − i x=\pm\sqrt{-1-i} . These are neither real nor purely imaginary.

(As an example, − 1 + i \sqrt{-1+i} has real part 2 4 cos ⁡ 3 π 8 \sqrt[4]{2}\cos\frac{3\pi}{8} and imaginary part 2 4 sin ⁡ 3 π 8 \sqrt[4]{2}\sin\frac{3\pi}{8} )

6 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...