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6)Find the solution set of the quadratic equation over the set of complex numbers.5x2 + 12x + 8 = 0A)x = −34 − i4 or −34 + i4B)x = −65 − 2i5 or −65 + 2i5C)x = −25i(11 − 2i) or −25i(11 + 2i)D)x = −16i(11 − 2i) or −16i(11 + 2i)

Sagot :

The given quadratic equation is expressed as

5x^2 + 12x + 8 = 0

The standard form of a quadratic equation is expressed as

ax^2 + bx + c = 0

By comparing both equations, we have

a = 5, b = 12, c = 8

We would solve for x by applying the general formula for solving quadratic equations which is expressed as

[tex]\begin{gathered} x\text{ = }\frac{-\text{ b }\pm\sqrt[]{b^2-4ac}}{2a} \\ By\text{ substituting the values, we have} \\ x\text{ = }\frac{-\text{ 12}\pm\sqrt[]{12^2-4(5\times8)}}{2\times5}\text{ } \\ x\text{ = }\frac{-\text{ 12 }\pm\sqrt[]{144\text{ - 160}}}{10}\text{ = }\frac{-\text{ 12}\pm\sqrt[]{-\text{ 16}}}{10} \\ x\text{ = }\frac{-\text{ 12}\pm4i}{10} \\ x\text{ = }\frac{-\text{ 12 + 4i}}{10}\text{ or x = }\frac{-\text{ 12 - 4i}}{10} \\ x\text{ = }\frac{-\text{ 12}}{10}\text{ + }\frac{4i}{10}\text{ or x = }\frac{-\text{ 12}}{10}\text{ - }\frac{4i}{10} \\ x\text{ = }\frac{-\text{ 6}}{5}+\text{ }\frac{2i}{5}\text{ or x = }\frac{-6}{5}-\text{ }\frac{2i}{5} \end{gathered}[/tex]

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