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3 years ago

What is the additive inverse of the complex number 9 – 4i?

Mathematics

1 answer:

stiks02 [169]3 years ago

8 0

Answer:-(9 - 4i) =

Step-by-step explanation:

Given the complex number 9 - 4i, the additive inverse is the number that when added to the original number, would equate to zero. Another term this may be known as is the additive identity. The additive inverse is simply:

(a + bi) = -(a + bi)

Given: 9 - 4i

-(9 - 4i) = -9 + 4i

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Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Given quadratic equation:

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Define the variables:

a = 5
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c = -2

Substitute the defined variables into the quadratic formula and solve for x:

$\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}$

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$\implies x=\dfrac{3 \pm \sqrt{19}}{5}$

Therefore, the exact solutions to the given quadratic equation are:

$x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}$

Learn more about the quadratic formula here:

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