Answer:
[tex]\sf x = \dfrac{-1 + i\sqrt{3} }{2} \quad or \quad \dfrac{-1 - i\sqrt{3} }{2}[/tex]
Explanation:
[tex]\sf Given : x^2 + x + 1 = 0[/tex]
Solve it using quadratic formula
[tex]\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \quad \:when\: \: ax^2 + bx + c = 0[/tex]
So, here constants are: a = 1, b = 1, c = 1
When the value's are substituted inside the formula
[tex]\sf \rightarrow x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(1)} }{2(1)}[/tex]
[tex]\sf \rightarrow x = \dfrac{-1 \pm \sqrt{-3} }{2}[/tex]
[tex]\sf \rightarrow x = \dfrac{-1 \pm i\sqrt{3} }{2}[/tex]
[tex]\sf \rightarrow x = \dfrac{-1 + i\sqrt{3} }{2} \quad or \quad \dfrac{-1 - i\sqrt{3} }{2}[/tex]
The answers are :
This problem can be solved using the quadratic equation.
x = -1 ± √1² - 4(1)(1) / 2a
x = -1 ± √1 - 4 / 2a
x = -1 ± √-3 / 2a
x = -1 ± i√3 / 2a (∴ i = √-1)
The possible values of x are :
