Faktor dari t2(kuadrat)-16t+22=.... Question from @arnindyaoc

September 2018 2 16 Report

Faktor dari t2(kuadrat)-16t+22=

Ganjar64 T12 = -b(+-)√(b²-4ac) / 2a
   = 16(+-)√(256-88) /2
   = 16(+-)√168 / 2
   = 16(+-)2√42 / 2
   = 8(+-)√42
t = 8+√42 atau t = 8-√42

0 votes Thanks 2

ibnnurjen T²-16t+22 = 0
kita cari akar-akar dengan rumus abc

$t_{1,2}= \frac{-b(+-) \sqrt{b^2-4ac} }{2a} \\ t_{1,2}= \frac{-(-16)(+-) \sqrt{(-16)^2-4(1)(22)} }{2(1)} \\ t_{1,2}= \frac{16(+-) \sqrt{256-88} }{2} \\ t_{1,2}= \frac{16(+-) \sqrt{168} }{2} \\ t_{1,2}= \frac{16(+-) \sqrt{168} }{2} \\ t_{1,2}= \frac{16(+-) 2\sqrt{42} }{2} \\ t_{1,2}=8(+-) \sqrt{42} \\ t_{1}=8+\sqrt{42} \\ atau \\ t_{2}=8-\sqrt{42}$

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