Question

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

Answer 1

Answer:

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

Step-by-step 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

Answer:

x = $\frac{12+\sqrt{138} }{6}$ and x = $\frac{12-\sqrt{138} }{6}$

Step-by-step explanation:

Let's use the quadratic formula, which states that for a quadratic of the form ax² + bc + c, the zeroes are: $x=\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $x=\frac{-b-\sqrt{b^2-4ac} }{2a}$.

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

$x=\frac{-b+\sqrt{b^2-4ac} }{2a}$

$x=\frac{-(-24)+\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24+\sqrt{552} }{12}=\frac{24+2\sqrt{138} }{12} =\frac{12+\sqrt{138} }{6}$

AND

$x=\frac{-b-\sqrt{b^2-4ac} }{2a}$

$x=\frac{-(-24)-\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24-\sqrt{552} }{12}=\frac{24-2\sqrt{138} }{12} =\frac{12-\sqrt{138} }{6}$

Thus, the zeroes are: x = $\frac{12+\sqrt{138} }{6}$ and x = $\frac{12-\sqrt{138} }{6}$.