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Find the solutions of the quadratic equation 3x^2 + 9x + 17 = 0.

 

 Apr 19, 2020
 #1
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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 ! cheeky

 Apr 19, 2020
edited by Guest  Apr 19, 2020
 #2
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+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:


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

.
 Apr 19, 2020

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