Rumus mp
Penurunan Rumus mp
Misalkan kita mempunyai fungsi kuadrat dengan akar x 1 x_1 x 1 x 2 x_2 x 2 x x x x p x_p x p
Jarak x 1 x_1 x 1 x p x_p x p x 2 x_2 x 2 x p x_p x p d d d Grafik Fungsi Kuadrat .
x 1 = x p − d x 2 = x p + d x 1 , 2 = x p ± d \begin{align*}
x_{1} & =x_{p}-d\\
x_{2} & =x_{p}+d\\
x_{1,2} & =x_{p}\pm d
\end{align*}
x 1 x 2 x 1 , 2 = x p − d = x p + d = x p ± d
Persamaan kuadrat dari grafik di atas bisa dibuat dengan:
( x − x 1 ) ( x − x 2 ) = 0 x 2 − x 2 x − x 1 x + x 1 x 2 = 0 x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0 \begin{align*}
(x-x_{1})(x-x_{2}) & =0\\
x^{2}-x_{2}x-x_{1}x+x_{1}x_{2} & =0\\
x^{2}-(x_{1}+x_{2})x+x_{1}x_{2} & =0\\
\end{align*}
( x − x 1 ) ( x − x 2 ) x 2 − x 2 x − x 1 x + x 1 x 2 x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0 = 0 = 0
Lihat Persamaan Kuadrat
Misalkan jumlah kedua akar adalah (sum ) s s s product ) p p p
x 1 + x 2 = s x 1 x 2 = p \begin{align*}
x_{1}+x_{2} &=s\\
x_1x_2 &= p
\end{align*}
x 1 + x 2 x 1 x 2 = s = p
Maka persamaan di atas bisa kita tulis lagi menjadi:
x 2 − s x + p = 0 x^{2}-sx+p =0
x 2 − s x + p = 0
Sekarang kita akan mencari rumus x 1 , 2 x_{1,2} x 1 , 2 s s s p p p
Kita tahu bahwa perkalian kedua akar adalah p p p
p = x 1 x 2 p = ( x p − d ) ( x p + d ) p = ( x p − d ) ( x p + d ) p = x p 2 − d 2 \begin{align*}
p &= x_1 x_2\\
p &= (x_{p}-d)(x_{p}+d)\\
p &= (x_p-d)(x_p+d)\\
p &={x_p}^{2}-d^{2}\\
\end{align*}
p p p p = x 1 x 2 = ( x p − d ) ( x p + d ) = ( x p − d ) ( x p + d ) = x p 2 − d 2
Dari situ kita bisa mengetahui nilai dari d d d
d 2 = x p 2 − p d = x p 2 − p \begin{align*}
d^{2} & ={x_p}^{2}-p\\
d &= \sqrt{{x_p}^{2}-p}
\end{align*}
d 2 d = x p 2 − p = x p 2 − p
Sekarang kita bisa memasukkan d d d x 1 x_1 x 1 x 2 x_2 x 2
x 1 , 2 = x p ± d = x p ± x p 2 − p \begin{align*}
x_{1,2} & =x_{p}\pm d \\
&=x_p\pm\sqrt{{x_p}^{2}-p}
\end{align*}
x 1 , 2 = x p ± d = x p ± x p 2 − p
Nilai x x x x p x_p x p
x p = x 1 + x 2 2 = s 2 \displaystyle{ x _{ p } = \frac{ x _{ 1 } + x _{ 2 } }{ 2 } = \frac{ s }{ 2 } } x p = 2 x 1 + x 2 = 2 s
Agar rumus lebih elegan kita misalkan m m m mean ) dari kedua akar. Bisa dituliskan
m = s 2 = x p \displaystyle{ m = \frac{ s }{ 2 } = x _{ p } } m = 2 s = x p
Kita bisa masukkan m m m
x 1 , 2 = x p ± ( x p ) 2 − p = m ± m 2 − p \displaystyle{ \begin{aligned}x _{ 1 , 2 } & = x _{ p } \pm \sqrt{ \left( x _{ p } \right) ^{ 2 } - p } \\ & = m \pm \sqrt{ m ^{ 2 } - p }\end{aligned} } x 1 , 2 = x p ± ( x p ) 2 − p = m ± m 2 − p
Kesimpulan Rumus mp
Pada persmaan x 2 − s x + p = 0 x^{2}-sx+p =0 x 2 − s x + p = 0
s = x 1 + x 2 m = x 1 + x 2 2 = s 2 p = x 1 x 2 x 1 , 2 = m ± m 2 − p x p = m \begin{align*}
s &= x_{1}+x_{2}\\
m &= \frac{x_{1}+x_{2}}{2}\\
&= \frac{s}{2} \\
p &= x_1x_2\\
x_{1,2} &=m\pm\sqrt{m^{2}-p}\\
x_p &= m \\
\end{align*}
s m p x 1 , 2 x p = x 1 + x 2 = 2 x 1 + x 2 = 2 s = x 1 x 2 = m ± m 2 − p = m
Rumus abc
Dari Rumus mp
Misalkan kita mempunyai persamaan berikut:
a x 2 + b x + c = 0 ax^2 + bx + c = 0
a x 2 + b x + c = 0
Rumus mp hanya bisa digunakan jika koefisien x 2 x^2 x 2 a a a
x 2 + b a x + c a = 0 x^2 + \frac{b}{a}x + \frac{c}{a} = 0
x 2 + a b x + a c = 0
Kita coba samakan dengan bentuk x 2 − s x + p = 0 x^2 - sx + p = 0 x 2 − s x + p = 0
x 2 + b a x + c a = 0 x 2 − ( − b a ) x + c a = 0 x 2 − s x + p = 0 \begin{align*}
x^{2}+\frac{b}{a}x+\frac{c}{a} & =0\\
x^{2}-\boxed{\left(-\frac{b}{a}\right)}x+\boxed{\frac{c}{a}} & =0\\
x^{2}-\qquad\boxed{s}\quad x+\boxed{p} & =0
\end{align*}
x 2 + a b x + a c x 2 − ( − a b ) x + a c x 2 − s x + p = 0 = 0 = 0
Maka berlaku:
s = − b a m = s 2 = − b 2 a p = c a \begin{align*}
s & =\frac{-b}{a}\\
m & =\frac{s}{2}\\
& =\frac{-b}{2a}\\
p & =\frac{c}{a}
\end{align*}
s m p = a − b = 2 s = 2 a − b = a c
Kita sekarang bisa menggunakan rumus mp:
x 1 , 2 = m ± m 2 − p = − b 2 a ± ( − b 2 a ) 2 − c a = − b 2 a ± b 2 4 a 2 − c a = − b 2 a ± b 2 4 a 2 − c a × 4 a 4 a = − b 2 a ± b 2 4 a 2 − 4 a c 4 a 2 = − b 2 a ± b 2 − 4 a c 4 a 2 = − b 2 a ± b 2 − 4 a c 2 a = − b ± b 2 − 4 a c 2 a \begin{align*}
x_{1,2} & =m\pm\sqrt{m^{2}-p}\\
& =\frac{-b}{2a}\pm\sqrt{\left(\frac{-b}{2a}\right)^{2}-\frac{c}{a}}\\
& =\frac{-b}{2a}\pm\sqrt{\frac{b^{2}}{4a^{2}}-\frac{c}{a}}\\
& =\frac{-b}{2a}\pm\sqrt{\frac{b^{2}}{4a^{2}}-\frac{c}{a}\times\frac{4a}{4a}}\\
& =\frac{-b}{2a}\pm\sqrt{\frac{b^{2}}{4a^{2}}-\frac{4ac}{4a^{2}}}\\
& =\frac{-b}{2a}\pm\sqrt{\frac{b^{2}-4ac}{4a^{2}}}\\
& =\frac{-b}{2a}\pm\frac{\sqrt{b^{2}-4ac}}{2a}\\
& =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}
\end{align*}
x 1 , 2 = m ± m 2 − p = 2 a − b ± ( 2 a − b ) 2 − a c = 2 a − b ± 4 a 2 b 2 − a c = 2 a − b ± 4 a 2 b 2 − a c × 4 a 4 a = 2 a − b ± 4 a 2 b 2 − 4 a 2 4 a c = 2 a − b ± 4 a 2 b 2 − 4 a c = 2 a − b ± 2 a b 2 − 4 a c = 2 a − b ± b 2 − 4 a c
Dengan Melengkapkan Kuadrat Sempurna
From Persamaan Kuadrat
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Prinsip utama: mengubah suatu persamaan menjadi p 2 + 2 p q + q 2 p^2 + 2pq + q^2 p 2 + 2 pq + q 2 ( p + q ) 2 (p + q)^2 ( p + q ) 2 ( p + q ) 2 = p + q \displaystyle{ \ \sqrt{ \left( p + q \right) ^{ 2 } } = p + q } ( p + q ) 2 = p + q
a x 2 + b x + c = 0 p 2 + 2 p q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}a & { \color{red} x ^{ 2 } } + b { \color{red} x } + c = 0 \\ & { \color{red} p ^{ 2 } } + 2 { \color{red} p } q + q ^{ 2 } = \left( p + q \right) ^{ 2 }\end{aligned} } a x 2 + b x + c = 0 p 2 + 2 p q + q 2 = ( p + q ) 2
Kita bisa misalkan x = p x = p x = p x 2 x^2 x 2 p 2 p^2 p 2 a a a
x 2 + b x a + c a = 0 p 2 + 2 p q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + \frac{ b { \color{red} x } }{ a } + \frac{ c }{ a } = 0 \\ & { \color{red} p ^{ 2 } } + 2 { \color{red} p } q + q ^{ 2 } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + a b x + a c = 0 p 2 + 2 p q + q 2 = ( p + q ) 2
Agar p p p x x x b x a \displaystyle{ \frac{ b x }{ a } } a b x 2 2 \displaystyle{ \frac{ 2 }{ 2 } } 2 2
x 2 + 2 b x 2 a + c a = 0 p 2 + 2 p q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + \frac{ { \color{red} 2 } b { \color{red} x } }{ 2 a } + \frac{ c }{ a } = 0 \\ & { \color{red} p ^{ 2 } } + { \color{red} 2 p } q + q ^{ 2 } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + 2 a 2 b x + a c = 0 p 2 + 2 p q + q 2 = ( p + q ) 2
Agar lebih mudah kita ubah posisi b b b x x x
x 2 + 2 x b 2 a + c a = 0 p 2 + 2 p q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + \frac{ { \color{red} 2 x } b }{ 2 a } + \frac{ c }{ a } = 0 \\ & { \color{red} p ^{ 2 } } + { \color{red} 2 p } q + q ^{ 2 } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + 2 a 2 x b + a c = 0 p 2 + 2 p q + q 2 = ( p + q ) 2
Agar lebih mudah lagi kita pisahkan bagian yang sudah kita samakan dengan yang belum.
x 2 + 2 x ⋅ b 2 a + c a = 0 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + { \color{red} 2 x } \cdot \frac{ b }{ 2 a } + \frac{ c }{ a } = 0 \\ & { \color{red} p ^{ 2 } } + { \color{red} 2 p } \cdot q + q ^{ 2 } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + 2 x ⋅ 2 a b + a c = 0 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2
Agar 2 x ⋅ b 2 a = 2 p ⋅ q \displaystyle{ 2 x \cdot \frac{ b }{ 2 a } = 2 p \cdot q } 2 x ⋅ 2 a b = 2 p ⋅ q b 2 a = q \displaystyle{ \frac{ b }{ 2 a } = q } 2 a b = q x = p x=p x = p
x 2 + 2 x ⋅ b 2 a + c a = 0 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + { \color{red} 2 x } \cdot { \color{blue} \frac{ b }{ 2 a } } + \frac{ c }{ a } = 0 \\ & { \color{red} p ^{ 2 } } + { \color{red} 2 p } \cdot { \color{blue} q } + q ^{ 2 } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + 2 x ⋅ 2 a b + a c = 0 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2
Sekarang kita perlu mencari yang sama dengan q 2 q^2 q 2 b 2 a = q \displaystyle{ \frac{ b }{ 2 a } = q } 2 a b = q q 2 = ( b 2 a ) 2 \displaystyle{ q ^{ 2 } = \left( \frac{ b }{ 2 a } \right) ^{ 2 } } q 2 = ( 2 a b ) 2 c a \displaystyle{ \frac{ c }{ a } } a c q 2 q^2 q 2 q 2 q^2 q 2 q 2 q^2 q 2
x 2 + 2 x ⋅ b 2 a + c a + q 2 = q 2 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + { \color{red} 2 x } \cdot { \color{blue} \frac{ b }{ 2 a } } + \frac{ c }{ a } + { \color{blue} q ^{ 2 } } = q ^{ 2 } \\ & { \color{red} p ^{ 2 } } + { \color{red} 2 p } \cdot { \color{blue} q } + { \color{blue} q ^{ 2 } } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + 2 x ⋅ 2 a b + a c + q 2 = q 2 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2
Karena q 2 = ( b 2 a ) 2 \displaystyle{ q ^{ 2 } = \left( \frac{ b }{ 2 a } \right) ^{ 2 } } q 2 = ( 2 a b ) 2
x 2 + 2 x ⋅ b 2 a + c a + ( b 2 a ) 2 = ( b 2 a ) 2 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + { \color{red} 2 x } \cdot { \color{blue} \frac{ b }{ 2 a } } + \frac{ c }{ a } + { \color{blue} \left( \frac{ b }{ 2 a } \right) ^{ 2 } } = \left( \frac{ b }{ 2 a } \right) ^{ 2 } \\ & { \color{red} p ^{ 2 } } + { \color{red} 2 p } \cdot { \color{blue} q } + { \color{blue} q ^{ 2 } } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + 2 x ⋅ 2 a b + a c + ( 2 a b ) 2 = ( 2 a b ) 2 p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2
Karena c a \displaystyle{ \frac{ c }{ a } } a c
x 2 + 2 x ⋅ b 2 a + ( b 2 a ) 2 = ( b 2 a ) 2 − c a p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2 \displaystyle{ \begin{aligned}& { \color{red} x ^{ 2 } } + { \color{red} 2 x } \cdot { \color{blue} \frac{ b }{ 2 a } } + { \color{blue} \left( \frac{ b }{ 2 a } \right) ^{ 2 } } = \left( \frac{ b }{ 2 a } \right) ^{ 2 } - \frac{ c }{ a } \\ & { \color{red} p ^{ 2 } } + { \color{red} 2 p } \cdot { \color{blue} q } + { \color{blue} q ^{ 2 } } = \left( p + q \right) ^{ 2 }\end{aligned} } x 2 + 2 x ⋅ 2 a b + ( 2 a b ) 2 = ( 2 a b ) 2 − a c p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2
Sekarang kita ubah x x x p p p b 2 a \displaystyle{ \frac{ b }{ 2 a } } 2 a b q q q
p 2 + 2 p ⋅ q + q 2 = ( b 2 a ) 2 − c a \displaystyle{ \begin{aligned}p ^{ 2 } + 2 p \cdot q + q ^{ 2 } & = \left( \frac{ b }{ 2 a } \right) ^{ 2 } - \frac{ c }{ a }\end{aligned} } p 2 + 2 p ⋅ q + q 2 = ( 2 a b ) 2 − a c
Karena p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2 \displaystyle{ p ^{ 2 } + 2 p \cdot q + q ^{ 2 } = \left( p + q \right) ^{ 2 } } p 2 + 2 p ⋅ q + q 2 = ( p + q ) 2
( p + q ) 2 = ( b 2 a ) 2 − c a \displaystyle{ \begin{aligned}\left( p + q \right) ^{ 2 } & = \left( \frac{ b }{ 2 a } \right) ^{ 2 } - \frac{ c }{ a }\end{aligned} } ( p + q ) 2 = ( 2 a b ) 2 − a c
Sekarang kita ubah kembali p p p x x x q q q b 2 a \displaystyle{ \frac{ b }{ 2 a } } 2 a b
( x + b 2 a ) 2 = ( b 2 a ) 2 − c a \displaystyle{ \begin{aligned}\left( x + \frac{ b }{ 2 a } \right) ^{ 2 } & = \left( \frac{ b }{ 2 a } \right) ^{ 2 } - \frac{ c }{ a }\end{aligned} } ( x + 2 a b ) 2 = ( 2 a b ) 2 − a c
Sekarang kita ambil akar dari kedua ruas:
( x + b 2 a ) 2 = ± ( b 2 a ) 2 − c a x + b 2 a = ± ( b 2 a ) 2 − c a \displaystyle{ \begin{aligned}\sqrt{ \left( x + \frac{ b }{ 2 a } \right) ^{ 2 } } & = \pm \sqrt{ \left( \frac{ b }{ 2 a } \right) ^{ 2 } - \frac{ c }{ a } } \\ x + \frac{ b }{ 2 a } & = \pm \sqrt{ \left( \frac{ b }{ 2 a } \right) ^{ 2 } - \frac{ c }{ a } }\end{aligned} } ( x + 2 a b ) 2 x + 2 a b = ± ( 2 a b ) 2 − a c = ± ( 2 a b ) 2 − a c
Karena kita membuat akar, maka salah satu ruas harus memiliki ± \displaystyle{ \pm } ±
Persamaan juga bisa ditulis seperti ini:
x + b 2 a = ± − c a + ( b 2 a ) 2 x + \frac{b}{2a} = \pm \sqrt{\frac{-c}{a} + (\frac{b}{2a})^2}
x + 2 a b = ± a − c + ( 2 a b ) 2
a x 2 + b x + c = 0 x + b 2 a = ± − c a + ( b 2 a ) 2 = ± − c a + b 2 4 a 2 = ± − c a × 4 a 4 a + b 2 4 a 2 = ± − 4 a c 4 a 2 + b 2 4 a 2 = ± b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 2 a x = − b 2 a ± b 2 − 4 a c 2 a x 1 , 2 = − b ± b 2 − 4 a c 2 a \begin{aligned}
ax^2 + bx + c &= 0\\
x+\frac{b}{2a} & =\pm \sqrt{\frac{-c}{a} +(\frac{b}{2a} )^{2}}\\
& =\pm \sqrt{\frac{-c}{a} +\frac{b^{2}}{4a^{2}}}\\
& =\pm \sqrt{\frac{-c}{a} \times \frac{4a}{4a} +\frac{b^{2}}{4a^{2}}}\\
& =\pm \sqrt{\frac{-4ac}{4a^{2}} +\frac{b^{2}}{4a^{2}}}\\
& =\pm \sqrt{\frac{b^{2} -4ac}{4a^{2}}}\\
x+\frac{b}{2a} & =\pm \frac{\sqrt{b^{2} -4ac}}{2a}\\
x & =\frac{-b}{2a} \pm \frac{\sqrt{b^{2} -4ac}}{2a}\\
x_{1,2} & =\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}
\end{aligned}
a x 2 + b x + c x + 2 a b x + 2 a b x x 1 , 2 = 0 = ± a − c + ( 2 a b ) 2 = ± a − c + 4 a 2 b 2 = ± a − c × 4 a 4 a + 4 a 2 b 2 = ± 4 a 2 − 4 a c + 4 a 2 b 2 = ± 4 a 2 b 2 − 4 a c = ± 2 a b 2 − 4 a c = 2 a − b ± 2 a b 2 − 4 a c = 2 a − b ± b 2 − 4 a c
Kesimpulan Rumus abc
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
