SenorMathWiz
Answered

Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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]

We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.