Question

What are the zeros of the quadratic function f(x) = 6x ^ 2 - 24x + 1 ?

Answer 1

x = 2 + sqrt(138)/6 , 2 - sqrt(138)/6

Explanation:

6x² - 24x + 1 = 0

x = [-(-24) +/- sqrt[24² - 4(6)(1)]/(2×6)

x = [24 +/- sqrt(552)]/12

x = 2 +/- sqrt(138)/6

Answer 2

x =
`(12+√(138) )/(6)` and x =
`(12-√(138) )/(6)`

Explanation:

Let's use the quadratic formula, which states that for a quadratic of the form ax² + bc + c, the zeroes are:
`x=(-b+√(b^2-4ac) )/(2a)` or
`x=(-b-√(b^2-4ac) )/(2a)`.

Here, a = 6, b = -24, and c = 1. Plug these in:

`x=(-b+√(b^2-4ac) )/(2a)`

`x=(-(-24)+√((-24)^2-4*6*1) )/(2*6)=(24+√(552) )/(12)=(24+2√(138) )/(12) =(12+√(138) )/(6)`

AND

`x=(-b-√(b^2-4ac) )/(2a)`

`x=(-(-24)-√((-24)^2-4*6*1) )/(2*6)=(24-√(552) )/(12)=(24-2√(138) )/(12) =(12-√(138) )/(6)`

Thus, the zeroes are: x =
`(12+√(138) )/(6)` and x =
`(12-√(138) )/(6)`.