Semua rumus di yang akan kita turunkan berlaku untuk persamaan:
a x 2 + b x + c = 0 ax^2 + bx + c = 0
a x 2 + b x + c = 0
Sehingga kita menggunakan rumus abc. Sebenarnya rumus jumlah akar dan perkalian bisa ditemukan
dengan cara pemisalan seperti saat kita mencari rumus mp dan abc.
From Rumus Kuadrat
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Pada persamaan a x 2 + b x + c = 0 ax^2 + bx + c =0 a x 2 + b x + c = 0
x 1 + x 2 = − b a x 1 x 2 = c a x p = − b 2 a x 1 , 2 = − b ± b 2 − 4 a c 2 a = − b ± D 2 a \begin{align*}
x_{1}+x_{2} & =\frac{-b}{a}\\
x_{1}x_{2} & =\frac{c}{a}\\
x_{p} & =\frac{-b}{2a}\\
x_{1,2} & =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\
& =\frac{-b\pm\sqrt{D}}{2a}
\end{align*}
x 1 + x 2 x 1 x 2 x p x 1 , 2 = a − b = a c = 2 a − b = 2 a − b ± b 2 − 4 a c = 2 a − b ± D
Rumus Jumlah Akar
x 1 + x 2 = − b − D 2 a + ( − b + D 2 a ) = − b − D − b + D 2 a = − 2 b 2 a = − b a \begin{aligned}
x_{1} +x_{2} & =\frac{-b-\sqrt{D}}{2a} +\left(\frac{-b+\sqrt{D}}{2a}\right)\\
& =\frac{-b-\sqrt{D} -b+\sqrt{D}}{2a}\\
& =\frac{-2b}{2a}\\
& =\frac{-b}{a}\\
\end{aligned}
x 1 + x 2 = 2 a − b − D + ( 2 a − b + D ) = 2 a − b − D − b + D = 2 a − 2 b = a − b
Rumus Perkalian Akar
x 1 × x 2 = − b − D 2 a × − b + D 2 a = b 2 − D 4 a 2 = b 2 − ( b 2 − 4 a c ) 4 a 2 = 4 a c 4 a 2 = c a \begin{aligned}
x_{1} \times x_{2} & =\frac{-b-\sqrt{D}}{2a} \times \frac{-b+\sqrt{D}}{2a}\\
& =\frac{b^{2} -D}{4a^{2}}\\
& =\frac{b^{2} -\left( b^{2} -4ac\right)}{4a^{2}}\\
& =\frac{4ac}{4a^{2}}\\
& =\frac{c}{a}
\end{aligned}
x 1 × x 2 = 2 a − b − D × 2 a − b + D = 4 a 2 b 2 − D = 4 a 2 b 2 − ( b 2 − 4 a c ) = 4 a 2 4 a c = a c
Rumus Pengurangan Akar
Jika x 1 < x 2 x_1 < x_2 x 1 < x 2
x 1 − x 2 = − b − D 2 a − ( − b + D 2 a ) = − b − D + b − D 2 a = − 2 D 2 a = − D a \begin{aligned}
x_{1} -x_{2} & =\frac{-b-\sqrt{D}}{2a} -\left(\frac{-b+\sqrt{D}}{2a}\right)\\
& =\frac{-b-\sqrt{D} +b-\sqrt{D}}{2a}\\
& =\frac{-2\sqrt{D}}{2a}\\
& =\frac{-\sqrt{D}}{a}
\end{aligned}
x 1 − x 2 = 2 a − b − D − ( 2 a − b + D ) = 2 a − b − D + b − D = 2 a − 2 D = a − D
Jika x 1 > x 2 x_1 > x_2 x 1 > x 2
x 1 − x 2 = − b + D 2 a − ( − b − D 2 a ) = − b + D + b + D 2 a = 2 D 2 a = D a \begin{aligned}
x_{1} -x_{2} & =\frac{-b+\sqrt{D}}{2a} -\left(\frac{-b-\sqrt{D}}{2a}\right)\\
& =\frac{-b+\sqrt{D} +b+\sqrt{D}}{2a}\\
& =\frac{2\sqrt{D}}{2a}\\
& =\frac{\sqrt{D}}{a}
\end{aligned}
x 1 − x 2 = 2 a − b + D − ( 2 a − b − D ) = 2 a − b + D + b + D = 2 a 2 D = a D
Rumus Penjumlahan Kuadrat Akar
x 1 2 + x 2 2 = ( − b − D 2 a ) 2 + ( − b + D 2 a ) 2 = b 2 + 2 b D + D 4 a 2 + b 2 − 2 b D + D 4 a 2 = 2 b 2 + 2 D 4 a 2 = b 2 + D 2 a 2 = b 2 + b 2 − 4 a c 2 a 2 = 2 b 2 − 4 a c 2 a 2 = b 2 − 2 a c a 2 = b 2 a 2 − 2 a c a 2 = b 2 a 2 − 2 c a = ( x 1 + x 2 ) 2 − 2 ( x 1 × x 2 ) \begin{aligned}
x_{1}^{2} +x_{2}^{2} & =\left(\frac{-b-\sqrt{D}}{2a}\right)^{2} +\left(\frac{-b+\sqrt{D}}{2a}\right)^{2}\\
& =\frac{b^{2} +2b\sqrt{D} +D}{4a^{2}} +\frac{b^{2} -2b\sqrt{D} +D}{4a^{2}}\\
& =\frac{2b^{2} +2D}{4a^{2}}\\
& =\frac{b^{2} +D}{2a^{2}}\\
& =\frac{b^{2} +b^{2} -4ac}{2a^{2}}\\
& =\frac{2b^{2} -4ac}{2a^{2}}\\
& =\frac{b^{2} -2ac}{a^{2}}\\
& =\frac{b^{2}}{a^{2}} -\frac{2ac}{a^{2}}\\
& =\frac{b^{2}}{a^{2}} -\frac{2c}{a}\\
& =( x_{1} +x_{2})^{2} -2( x_{1} \times x_{2})
\end{aligned}
x 1 2 + x 2 2 = ( 2 a − b − D ) 2 + ( 2 a − b + D ) 2 = 4 a 2 b 2 + 2 b D + D + 4 a 2 b 2 − 2 b D + D = 4 a 2 2 b 2 + 2 D = 2 a 2 b 2 + D = 2 a 2 b 2 + b 2 − 4 a c = 2 a 2 2 b 2 − 4 a c = a 2 b 2 − 2 a c = a 2 b 2 − a 2 2 a c = a 2 b 2 − a 2 c = ( x 1 + x 2 ) 2 − 2 ( x 1 × x 2 )
Rumus Penjumlahan Kuadrat Akar
x 1 3 + x 2 3 = ( − b − D 2 a ) 3 + ( − b + D 2 a ) 3 = ( − b 3 − 3 b 2 D − 3 b D − D D ) 8 a 3 + ( − b 3 + 3 b 2 D − 3 b D + D D 8 s 2 ) = − 2 b 3 − 6 b D 8 a 3 = − b 3 − 3 b D 4 a 3 = − b 3 − 3 b ( b 2 − 4 a c ) 4 a 3 = − b 3 − 3 b 3 + 12 a b c 4 a 3 = − 4 b 3 + 12 a b c 4 a 3 = − 4 b 3 4 a 3 + 12 a b c 4 a 3 = ( − b a ) 3 + 3 b c a 2 = ( − b a ) 3 + 3 × b a × c a = ( x 1 + x 2 ) 3 + 3 ( x 1 + x 2 ) ( x 1 × x 2 ) \begin{aligned}
x_{1}^{3} +x_{2}^{3} & =\left(\frac{-b-\sqrt{D}}{2a}\right)^{3} +\left(\frac{-b+\sqrt{D}}{2a}\right)^{3}\\
& =\frac{\left( -b^{3} -3b^{2}\sqrt{D} -3bD-D\sqrt{D}\right)}{8a^{3}} +\left(\frac{-b^{3} +3b^{2}\sqrt{D} -3bD+D\sqrt{D}}{8s^{2}}\right)\\
& =\frac{-2b^{3} -6bD}{8a^{3}}\\
& =\frac{-b^{3} -3bD}{4a^{3}}\\
& =\frac{-b^{3} -3b\left( b^{2} -4ac\right)}{4a^{3}}\\
& =\frac{-b^{3} -3b^{3} +12abc}{4a^{3}}\\
& =\frac{-4b^{3} +12abc}{4a^{3}}\\
& =\frac{-4b^{3}}{4a^{3}} +\frac{12abc}{4a^{3}}\\
& =\left(\frac{-b}{a}\right)^{3} +\frac{3bc}{a^{2}}\\
& =\left(\frac{-b}{a}\right)^{3} +3\times \frac{b}{a} \times \frac{c}{a}\\
& =( x_{1} +x_{2})^{3} +3( x_{1} +x_{2})( x_{1} \times x_{2})
\end{aligned}
x 1 3 + x 2 3 = ( 2 a − b − D ) 3 + ( 2 a − b + D ) 3 = 8 a 3 ( − b 3 − 3 b 2 D − 3 b D − D D ) + ( 8 s 2 − b 3 + 3 b 2 D − 3 b D + D D ) = 8 a 3 − 2 b 3 − 6 b D = 4 a 3 − b 3 − 3 b D = 4 a 3 − b 3 − 3 b ( b 2 − 4 a c ) = 4 a 3 − b 3 − 3 b 3 + 12 ab c = 4 a 3 − 4 b 3 + 12 ab c = 4 a 3 − 4 b 3 + 4 a 3 12 ab c = ( a − b ) 3 + a 2 3 b c = ( a − b ) 3 + 3 × a b × a c = ( x 1 + x 2 ) 3 + 3 ( x 1 + x 2 ) ( x 1 × x 2 )
Kesimpulan
Untuk persmaan a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0
x 1 + x 2 = − b a x 1 x 2 = c a x 1 − x 2 = − D a ( x 1 < x 2 ) x 1 − x 2 = D a ( x 1 > x 2 ) x 1 2 + x 2 2 = ( x 1 + x 2 ) 2 − 2 ( x 1 × x 2 ) x 1 3 + x 2 2 = ( x 1 + x 2 ) 3 + 3 ( x 1 + x 2 ) ( x 1 × x 2 ) \begin{align*}
x_{1}+x_{2} & =\frac{-b}{a}\\
x_{1}x_{2} & =\frac{c}{a}\\
x_{1}-x_{2} & =\frac{-\sqrt{D}}{a} & (x_{1}<x_{2})\\
x_{1}-x_{2} & =\frac{\sqrt{D}}{a} & (x_{1}>x_{2})\\
x_{1}^{2}+x_{2}^{2} & =(x_{1}+x_{2})^{2}-2(x_{1}\times x_{2})\\
x_{1}^{3}+x_{2}^{2} & =(x_{1}+x_{2})^{3}+3(x_{1}+x_{2})(x_{1}\times x_{2})
\end{align*}
x 1 + x 2 x 1 x 2 x 1 − x 2 x 1 − x 2 x 1 2 + x 2 2 x 1 3 + x 2 2 = a − b = a c = a − D = a D = ( x 1 + x 2 ) 2 − 2 ( x 1 × x 2 ) = ( x 1 + x 2 ) 3 + 3 ( x 1 + x 2 ) ( x 1 × x 2 ) ( x 1 < x 2 ) ( x 1 > x 2 )