Hasil akar_akar persamaan mudarat 2x²_ 4x + 6 = 0 adalah.... Question from @noventresandya

October 2018 2 20 Report

Hasil akar_akar persamaan mudarat 2x²_ 4x + 6 = 0 adalah.......

DrSpongebob $2x^2-4x+6=0\\x_1= \frac{-(-4)+ \sqrt{(-4)^2-4(2)(6)} }{2(2)}= \frac{4+ \sqrt{-32} }{4}= \frac{4+ 4\sqrt{-2} }{4}= 1+ \sqrt{-2}\\x_2=1- \sqrt{-2}$

Semoga membantu...

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noventresandya mkasih

kardillah 2 $x^{2}$ - 4x + 6 = 0
Dik : a = 2
b = -4
c = 6
Dicari $\\ x_{1}$ atau $x_{2}$ ?
Penyelesaiannya :
$x_{1}$ = $\frac{-b + \sqrt{ b^{2} - 4ac } }{2a}$
$x_{1}$ = $\frac{-(-4) + \sqrt{ -4^{2} - 4.2.6 } }{2.2}$
$x_{1}$ = $\frac{ 4 + \sqrt{16 - 48 } }{4}$
$x_{1}$ = $\frac{ 4 + \sqrt{- 32 } }{4}$
$x_{1}$ = $\frac{4 + 4 \sqrt{-2} }{4}$
$x_{1}$ = $1 + \sqrt{-2}$

ATAU

$x_{2}$ = $\frac{-b - \sqrt{ b^{2} - 4ac } }{2a}$
$x_{2}$ = $\frac{-(-4) - \sqrt{ -4^{2} - 4.2.6 } }{2.2}$
$x_{2}$ = $\frac{ 4 - \sqrt{16 - 48 } }{4}$
$x_{2}$ = $\frac{ 4 - \sqrt{- 32 } }{4}$
$x_{2}$ = $\frac{4 - 4 \sqrt{-2} }{4}$
$x_{2}$ = $1 - \sqrt{-2}$

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