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Solve each question with the quadratic formula
6a² + 10 = 0


I know the answer but I'm not to sure how imaginary roots work. I would really appreciate if someone could explain how imaginary roots work! :)

Sagot :

Answer:

[tex]\boxed{\sf a=\cfrac{\sqrt{15i} }{3}}[/tex]

[tex]\boxed{\sf a=-\cfrac{\sqrt{15i} }{3}}[/tex]

Step-by-step explanation:

[tex]\sf 6a^2 + 10 = 0[/tex]

We'll solve this equation using The Quadratic Formula:

[tex]\boxed{\sf \frac{-b\pm \sqrt{b^2-4ac} }{2a} }[/tex]

[tex]\sf 6a^2 + 10 = 0[/tex]

Substitute 6: a, 0: b, 10: c

[tex]\longmapsto\sf a=\cfrac{0\pm\sqrt{0^2-4\times 6\times 10} }{2\times 6}[/tex]

→ Square 0

→ Multiply -4 * 6= -24

→ Multiply -24 * 10 = -240

[tex]\longmapsto\sf a=\cfrac{0\pm\sqrt{-240} }{2\times 6}[/tex]

→ Take the square root of -240:

→ Multiply 2 * 6= 12

Now, Separate solutions:

[tex]\longmapsto\sf a=\cfrac{0\pm 4\sqrt{15i} }{12}[/tex]

Now, Separate solutions:

First, solve the equation when ± is plus:

[tex]\longmapsto \boxed{\sf a=\cfrac{\sqrt{15i} }{3}}[/tex]

___

Now, solve it when ± is minus:

[tex]\longmapsto \boxed{\sf a=-\cfrac{\sqrt{15i} }{3}}[/tex]

__________________________________________

[tex]\sf\longmapsto 6a^2+10=0[/tex]

  • a=6
  • b=0
  • c=10

[tex]\sf\longmapsto a=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex](This a is root of equation)

[tex]\sf\longmapsto a=\dfrac{-0\pm\sqrt{0^2-4(6)(10)}}{2(6)}[/tex]

[tex]\sf\longmapsto a=\dfrac{\pm\sqrt{-240}}{12}[/tex]

[tex]\sf\longmapsto a=\dfrac{\pm\sqrt{240}i}{12}[/tex]

[tex]\sf\longmapsto a=\dfrac{\pm 4√15i}{12}[/tex]

[tex]\sf\longmapsto a=\dfrac{\pm √15i}{3}[/tex]

[tex]\sf\longmapsto a=\dfrac{\pm √3√5i}{3}[/tex]

[tex]\sf\longmapsto a=\pm\dfrac{√5i}{\sqrt{3}}[/tex]

[tex]\sf\longmapsto a=\pm\sqrt{\dfrac{5}{3}}i [/tex]

What is i?

i stands for iota.

[tex]\sf\longmapsto i=\sqrt{-1}[/tex]

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