#1**+2 **

Use Quadratic Formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) a = 3 b = 9 c= 17 results in - 3/2 +-............ You can do it !

ElectricPavlov Apr 19, 2020

edited by
Guest
Apr 19, 2020

#2**+2 **

Trying factoring, This is unfactorable, as 3 * 17 = 51, and since 17 and 3 are both prime, this means that it is impossible to find other ways to split up 51 other than 17 3, and 51, 1 so it is unfactorable.

Using the quadratic equation with a = 3, b = 9, c = 17.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Plugging in a b and c for a = 3, b = 9, c = 17,

\(x = {-9 \pm \sqrt{9^2-4(3)(17)} \over 2(3)}\)

This simplifies too:

\(x = {-9 \pm \sqrt{81-4(51)} \over 6}\)

Normally, since 81 - 4(51) < 0, we would say unsolvable and move on, but since complex numbers are allowed, we can continue.

\(x = {-9 \pm \sqrt{81-204} \over 6}\) = \(x = {-9 \pm \sqrt{-123} \over 6}\) = \(x = {-9 \pm (\sqrt{123})(\sqrt{-1}) \over 6}\) i = sqrt(-1), so \(x = {-9 \pm \sqrt{123}i \over 6}\)

Simplifying this to a complex number,

\(x = {\frac{-9}{6}} \pm {\sqrt{123}i \over 6} \), and simplifying the fractions, the final answer is:

\(x = {-\frac{2}{3}} \pm {\sqrt{123} \over 6 }i \)

shad0w Apr 19, 2020