Quadratic Function with Imaginary Coefficients

Algebra Level 2

Find the zeroes of the quadratic function above. To see the solution for this problem, find a post named "Solutions for Quadratic Function with Imaginary Coefficients".

-1.5i and 4i -1.4i and 3i -1.4i and 4i -1.5i and 3i

Luis Tabule by Luis Tabule , 23, Philippines

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2 solutions

2 i x 2 + 5 x + 12 i = 0 2ix^2+5x+12i=0

x = − 5 ± 5 2 − 4 ( 2 i ) ( 12 i ) 4 i = − 5 ± 121 4 i = − 5 ± 11 4 i x = \dfrac {-5\pm \sqrt{5^2-4(2i)(12i)}}{4i} = \dfrac {-5\pm \sqrt{121}}{4i} = \dfrac {-5\pm 11}{4i}

⇒ x = { 6 4 i = 1.5 i = 1.5 i i 2 = 1.5 i − 1 = − 1.5 i − 16 4 i = − 4 i = − 4 i i 2 = − 4 i − 1 = 4 i \Rightarrow x = \begin{cases} \dfrac {6}{4i} = \dfrac {1.5}{i} = \dfrac {1.5i}{i^2} = \dfrac {1.5i}{-1} = - 1.5i \\ \dfrac {-16}{4i} = \dfrac {-4}{i} = \dfrac {-4i}{i^2} = \dfrac {-4i}{-1} = 4i \end{cases}

Therefore, the roots are -1.5i and 4i \boxed {\text{-1.5i and 4i}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Mhar Ariz Marino
Dec 8, 2014

by looking at the choices. the product of the roots is 6 then the roots should have a product of 6.

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