Answered

What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1?

x = –2 + StartRoot StartFraction 1 Over 2 EndFraction EndRoot or x = –2 – StartRoot StartFraction 1 Over 2 EndFraction EndRoot
x = 2 + StartRoot StartFraction 1 Over 2 EndFraction EndRoot or x = 2 – StartRoot StartFraction 1 Over 2 EndFraction EndRoot
x = –2 + StartRoot StartFraction 23 Over 6 EndFraction EndRoot or x = –2 – StartRoot StartFraction 23 Over 6 EndFraction EndRoot
x = 2 + StartRoot StartFraction 23 Over 6 EndFraction EndRoot or x = 2 – StartRoot StartFraction 23 Over 6 EndFraction EndRoot

Sagot :

Answer:

[tex]x=\frac{12+\sqrt{138}}{6}\:\:or\:\:x=\frac{12-\sqrt{138}}{6}[/tex]

Step-by-step explanation:

[tex]f(x)=6x^2-24x+1\\\\0=6x^2-24x+1\\\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-(-24)\pm\sqrt{(-24)^2-4(6)(1)}}{2(6)}\\ \\x=\frac{24\pm\sqrt{576-24}}{12}\\\\x=\frac{24\pm2\sqrt{138}}{12}\\\\x=\frac{12\pm\sqrt{138}}{6}[/tex]

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