Value of x is: [tex]x=6\pm\sqrt{23}i[/tex]
Option B is correct option.
Step-by-step explanation:
We need to solve the equation [tex]x^2-12x+59=0[/tex] and find value of x.
Using quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Putting values of a, b and c and finding the value of x
a=1, b=-12 and c=59
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(59)}}{2(1)}\\x=\frac{12\pm\sqrt{144-236}}{2}\\x=\frac{12\pm\sqrt{-92}}{2}\\x=\frac{12\pm\sqrt{92}i}{2}\\We\,\,know\,\sqrt{-1} \,\,is\,\,i\\x=\frac{12\pm2\sqrt{23}i}{2}\\x=\frac{2(6\pm\sqrt{23}i)}{2}\\x=6\pm\sqrt{23}i[/tex]
So, value of x is: [tex]x=6\pm\sqrt{23}i[/tex]
Option B is correct option.
