Jika persamaan kuadrat x pangkat 2 - 2x + 3 = 0 Mempunyai akae-akar alpha dan beta, maka persamaan k.... Question from @ZiendyLestuny

October 2018 1 26 Report

Jika persamaan kuadrat x pangkat 2 - 2x + 3 = 0
Mempunyai akae-akar alpha dan beta, maka persamaan kudrat baru yg akar2nya (alpha kuadrat + beta kudrat) dan ( 1/alpha kuadrat + 1/beta kudrat) adalah...

MicoArrafi $\bigstar\ \bold{\underline{Solution\ best\ answer}}\ \bigstar$

$$\begin{align} x^2-2x+3&=0\\\\\alpha+\beta&=-\frac{b}{a}=-\frac{-2}{1}=2\\\alpha\times\beta&=\frac{c}{a}=\frac{3}{1}=3 \end{align}$

$$\begin{align} \left(\alpha^2+\beta^2\right)+\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)&=\left((\alpha+\beta)^2-2\alpha\beta\right)+\left(\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}\right)\\&=\left((\alpha+\beta)^2-2\alpha\beta\right)+\left(\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}\right)\\&=\left(2^2-2\times3\right)+\left(\frac{2^2-2\times3}{3^2}\right)\\&=-2+\frac{-2}{9}\\&=-\frac{20}{9} \end{align}$

$$\begin{align} \left(\alpha^2+\beta^2\right)\times\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right)&=\left((\alpha+\beta)^2-2\alpha\beta\right)\times\left(\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}\right)\\&=\left((\alpha+\beta)^2-2\alpha\beta\right)\times\left(\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}\right)\\&=\left(2^2-2\times3\right)\times\left(\frac{2^2-2\times3}{3^2}\right)\\&=-2\times\frac{-2}{9}\\&=\frac{4}{9} \end{align}$

$\bold{Persamaan\ kuadrat\ baru\ :}$

$\boxed{x^2+\frac{20}{9}x+\frac{4}{9}=0}$ atau $\boxed{9x^2+20x+4=0}$

1 votes Thanks 1

ZiendyLestuny terima kasih banyak...:D

MicoArrafi Iya, sama-sama :D

Tulislah reaksi peruraian dari larutan : a) Fe2 (SO4)3 b) Ba (No3)2

Jika diketahui segitiga PQR siku-siku di Q dan sin P = 1/3 akar 5, maka cos P adalah...

Jika diketahui sim alpha = t dan alpha sudut tumpul, maka nilai tan alpha adalah...

Jika diketahui sin alpha = t dan alpha sudut tumpul, maka nilai tan alpha adalah...

Recommend Questions

Life Enjoy

Get in touch

Social