Answer:
Solving the expression [tex]3x^2-2x=-1[/tex] we get: [tex]\mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}[/tex]
Step-by-step explanation:
We need to solve the expression: [tex]3x^2-2x=-1[/tex]
This is a quadratic expression and it can be solved using quadratic formula
Solving:
[tex]3x^2-2x=-1\\[/tex]
we can write it as:
[tex]3x^2-2x+1=0[/tex]
The quadratic formula is: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
where a = 3, b = -2 and c= 1
Putting values and solving:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(1)}}{2(3)}\\x=\frac{2\pm\sqrt{4-12}}{2(3)}\\x=\frac{2\pm\sqrt{-8}}{6}\\We\:know\:that\:\sqrt{-1}=i\\x=\frac{2\pm\sqrt{8}\sqrt{-1} }{6} \\We\:know\:\sqrt{8}=\sqrt{2\times 2 \times 2}=\sqrt{2^2 \times 2}=2\sqrt{2} \\x=\frac{2\pm2\sqrt{2}i }{6}\\Now,\\x=\frac{2+2\sqrt{2}i }{6}\:or\:x=\frac{2-2\sqrt{2}i }{6}\\x=\frac{2(1+\sqrt{2}i) }{6}\:or\:x=\frac{2(1-\sqrt{2}i) }{6}\\x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}[/tex]
So, solving the expression [tex]3x^2-2x=-1[/tex] we get: [tex]\mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}[/tex]
