Step-by-step explanation:
[tex]x^{2} - 12x + 59 = 0[/tex]
Given equaiton is in the form of ax^2 +bx+c=0
we apply quadratic formula to solve for x
[tex]x= \frac{-b+-\sqrt{b^2-4ac} }{2a}[/tex]
a= 1 b = -12 and c= 59
[tex]x= \frac{12+-\sqrt{(-12)^2-4(1)(59)}}{2(1)}[/tex]
[tex]x= \frac{12+-\sqrt{92}}{2}[/tex]
[tex]x= \frac{12+-2\sqrt{23}}{2}[/tex]
Divide the 12 and square root terms by 2
[tex]x=6+-\sqrt{23}[/tex]
so [tex]x=6+\sqrt{23}[/tex] and [tex]x=6-\sqrt{23}[/tex]
Answer:
Solution for x:
[tex]6+\sqrt{23}i[/tex] and [tex]6-\sqrt{23}i[/tex]
Step-by-step explanation:
Given: [tex]x^2-12x+59=0[/tex]
It is quadratic equation and to solve for x.
using quadratic formula:
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
where, a=1, b=-12 and c=59
Put the values into the formula
[tex]x=\dfrac{12\pm\sqrt{12^2-4\cdot 1\cdot -59}}{2(1)}[/tex]
[tex]x=\dfrac{12\pm \sqrt{-92}}{2}[/tex]
As we know, [tex]\sqrt{-1}=i[/tex]
[tex]x=6\pm\sqrt{23}i[/tex]
Exact value of x : [tex]6+\sqrt{23}i[/tex] and [tex]6-\sqrt{23}i[/tex]
