Persamaan dengan pangkat tertinggi dua.
Bentuk:
a x 2 + b x + c = 0 ax^2 + bx + c = 0
a x 2 + b x + c = 0
Dengan a, b, dan c ∈ ℜ \in \Re ∈ ℜ a ≠ 0 a \ne 0 a = 0
Bentuk-Bentuk Persamaan Kuadrat
Persamaan Kuadrat Real, jika a, b dan c ∈ ℜ \in \Re ∈ ℜ
Persamaan Kuadrat Rasioanal, jika a, b, dan c bilangan rasional
Persamaan Kuadrat Tak Lengkap, jika c = 0 c = 0 c = 0
Persamaan Kuadrat Mutlak Sejati, jika b = 0 b=0 b = 0
Menyelesaikan Persamaan Kuadrat
Pemfaktoran
Prinsip utama pemfaktoran: Jika x ⋅ y = 0 x\cdot y = 0 x ⋅ y = 0 x = 0 x = 0 x = 0 y = 0 y = 0 y = 0
Untuk jenis a x 2 + b x = 0 ax^2 + bx = 0 a x 2 + b x = 0
a x 2 + b x = 0 x ( a x + b ) = 0 \begin{aligned}
ax^2 + bx &= 0 \\
x(ax + b) &= 0
\end{aligned}
a x 2 + b x x ( a x + b ) = 0 = 0
Jadi x = 0 x = 0 x = 0 a x + b = 0 ax + b = 0 a x + b = 0
Untuk jenis a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0
Carilah dua angka (r dan s) yang jika dikalikan menghasilkan a.
r ⋅ s = a r\cdot s = a
r ⋅ s = a
Carilah dua angka (p dan q) yang jika dikalikan menghasilkan c dan jika rp ditambah sq
menghasilkan b.
p ⋅ q = c r p + s q = b \begin{aligned}
p\cdot q &= c \\
rp + sq &= b
\end{aligned}
p ⋅ q r p + s q = c = b
Kemudian tuliskan sebagai:
( r x + q ) ( s x + p ) = 0 r ⋅ s = a p ⋅ q = c r p + s q = b \begin{aligned}
(rx + q)(sx + p) &= 0 \\
r\cdot s &= a \\
p\cdot q &= c \\
rp + sq &= b \\
\end{aligned}
( r x + q ) ( s x + p ) r ⋅ s p ⋅ q r p + s q = 0 = a = c = b
Melengkapkan Kuadrat Sempurna
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
Rumus Kuadrat
Rumus mp
From Rumus Kuadrat
Go to text
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
From Rumus Kuadrat
Go to text
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
Diskriminan
Formula di bawah akar pada rumus abc disebut dengan diskriminan.
D = b 2 − 4 a c D = b^2 - 4ac
D = b 2 − 4 a c
Nilai dari D menentukan jenis akar-akar:
D > 0 D > 0 D > 0 D \sqrt{D} D
D kuadeat sempurna, dua akarnya rasional
D kuadrat tidak sempurna, dua akarnya irasional
D = 0 D = 0 D = 0 D = 0 \sqrt{D} = 0 D = 0 D < 0 D < 0 D < 0 D \sqrt{D} D
Hasil Penjumlahan dan Perkalian Akar
From Rumus Jumlah dan Perkalian Akar
Go to text
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 )
Menyusun Persamaan Kuadrat
Jika Diketahui Akar-Akarnya
( x + p ) ( x + q ) = 0 x 2 + p x + q x + p q = 0 \begin{aligned}
(x+p)(x+q) &= 0 \\
x^2+px+qx+pq &=0
\end{aligned}
( x + p ) ( x + q ) x 2 + p x + q x + pq = 0 = 0
Akar-akarnya:
x + p = 0 x 1 = − p p = − x 1 x + q = 0 x 2 = − q q = − x 2 \begin{aligned}
x+p &= 0 \\
x_1 &= -p & \\
p &= -x_1 \\
x+q &= 0 \\
x_2 &= -q \\
q &= -x_2
\end{aligned}
x + p x 1 p x + q x 2 q = 0 = − p = − x 1 = 0 = − q = − x 2
Jadi:
( x − x 1 ) ( x − x 2 ) = 0 (x-x_1)(x-x_2) = 0
( x − x 1 ) ( x − x 2 ) = 0
Jika Diketahui Hasil Jumlah dan Kali Kedua Akarnya
( x + p ) ( x + q ) = 0 x 2 + ( p + q ) x + p q = 0 \begin{aligned}
(x+p)(x+q) &= 0 \\
x^2+(p+q)x+pq &=0 \\
\end{aligned}
( x + p ) ( x + q ) x 2 + ( p + q ) x + pq = 0 = 0
Jumlah dan kali kedua akar:
p + q = − x 1 + ( − x 2 ) = − ( x 1 + x 2 ) p q = ( − x 1 ) ( − x 2 ) = x 1 x 2 \begin{aligned}
p + q &= -x_1 + (-x_2) \\
&= -(x_1+x_2) \\
pq &= (-x_1)(-x_2) \\
&= x_1x_2
\end{aligned}
p + q pq = − x 1 + ( − x 2 ) = − ( x 1 + x 2 ) = ( − x 1 ) ( − x 2 ) = x 1 x 2
Jadi:
x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0 x^2 - (x_1+x_2)x + x_1x_2 = 0
x 2 − ( x 1 + x 2 ) x + x 1 x 2 = 0
Jika Diketahui Hubungan Dengan Akar Persamaan Lainnya
Cari x 1 + x 2 x_1 + x_2 x 1 + x 2 x 1 x 2 x_1x_2 x 1 x 2