Question

0 = 5² - 2х + 6

a) x = 1 ± 3i / 2
b) x = 1 ± √11 / 5
c) x = 1 ± i √29 / 5​

Answer 1

c) x = 1 ± i √29 / 5​

Explanation:

assuming we are solving for x we can use the quadratic formula

quadratic formula :
`(-b+or-√(b^2-4(a)(c)) )/(2(a))`

where the values of a,b and c are derived from the equation

the equation is put in quadratic form : ax² + bx + c

so in 5x² -2x + 6 a = 5 , b = - 2 and c = 6

we then plug these values of a, b and c into the quadratic formula

recall quadratic formula
`(-b+or-√(b^2-4(a)(c)) )/(2(a))`

a = 5 , b = - 2 , c = 6

`(-(-2)+or-√((-2)^2-4(5)(6)) )/(2(5))`

remove parenthesis at -(-2) to get positive 2 because the negative signs cancel out

`(2+or-√((-2)^2-4(5)(6)) )/(2(5))`

evaluate exponent

`(2+or-√(4-4(5)(6)) )/(2(5))`

multiply -4(5)(6)

`(2+or-√(4-120) )/(2(5))`

subtract 120 from 4

`(2+or-√(-116) )/(2(5))`

multiply 2 and 5

`(2+or-√(-116) )/(10)`

because we cant have a negative square root we replace the negative sign with i (which equals -1 ) and add it to the outside of the square root

`(2+or-i√(116) )/(10)`

we then simplify the radical

`(2+or-2i√(29) )/(10)`

we then simplify the fraction

2/10 = 1/5 so both 2's cancel out

`(1+or-i√(29) )/(10)`

the answer is C