- bentuk
- ax2+bx+c=0
Nilai hasil akar terdiri dari tiga jenis yaitu memfaktorkan, pengkuadratan serta rumus ABC.
contoh
- tentukan nilai akar dari persamaan x2-16x+55=0!
- cara 1
Jawaban
- cara 2
Jawaban
- cara 3
Jawaban
bentuk:
- ax2+bx+c=0
- x2+b/ax+c/a=0
- dengan menggunakan (x-x1)(x-x2)
- (x-x1)(x-x2)=0
- x2-(x1+x2)x+x1x2=0
- x2-(-b/a)x+c/a=0
![{\displaystyle x_{1}+x_{2}=-{\frac {b}{a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62332ad46f539c2a729aed1059d4b5f1cb0c4b1a)
![{\displaystyle x_{1}\cdot x_{2}={\frac {c}{a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17e7a76d71f5c8b24a654e2227e7d4e628eeb609)
contoh
- tentukan nilai p dari persamaan x2-8x+p=0 dimana salah satu akarnya 2 lebih dari akar lainnya!
Jawaban
- bentuk
- x' = x diubah menjadi x = x' dengan menggunakan sifat akar.
Persamaan kuadrat baru
Pernyataan |
Akar lama |
Akar baru |
Persamaan kuadrat baru
|
lebihnya dari |
x'=x+p |
x=x'-p |
a(x'-p)2+b(x'-p)+c=0
|
kurangnya dari |
x'=x-p |
x=x'+p |
a(x'+p)2+b(x'+p)+c=0
|
kalinya dari |
x'=px |
x=x'/p |
a(x')2+bpx'+cp2=0
|
baginya dari |
x'=x/p |
x=px' |
ap2(x')2+bpx'+c=0
|
berlawanan |
x'=-x |
x=-x' |
a(x')2-bx'+c=0
|
kebalikan |
x'=1/x |
x=1/x' |
c(x')2+bx'+a=0
|
kuadratnya |
x=(x')2 |
![{\displaystyle x'={\sqrt {x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b6feff08e818f0dd55501ec027c6acdaffbe1cd) |
a2(x')2-(b2-2ac)x'+c2=0
|
akarnya |
![{\displaystyle x={\sqrt {x'}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32444a6c8f93dbf533ae034afe5c67714892bcc7) |
(x')=x2 |
a(x')4-b(x')2+c=0
|
contoh
- tentukan persamaan kuadrat baru dari 2x2-3x+1=0 yang akar-akarnya p-2 dan q-2!
Jawaban
- tentukan persamaan kuadrat baru dari x2-x+3=0 yang akar-akarnya pq dan p+q!
Jawaban
- tentukan persamaan kuadrat baru dari 5x2+2x-1=0 yang akar-akarnya 1/q dan 1/q!
Jawaban
Diskriminan dan kriteria akar-akar
[sunting]
- Diskriminan (D) = b2-4ac
Kriteria akar-akar
Pernyataan |
Kriteria
|
Kedua akar riil yang berbeda (D>0)
|
bertanda positif |
x1+x2>0 dan x1x2>0
|
bertanda negatif |
x1+x2<0 dan x1x2>0
|
berlawanan |
x1x2<0
|
Akar riil yang sama (D=0)
|
berlawanan |
b=0
|
kebalikan |
c=a
|
Akar imajiner (D<0)
|
contoh
- tentukan nilai b yang memenuhi persamaan x2+(b-8)x+(b+3)=0 yang memiliki kedua akar yang berbeda dan bertanda positif!
Jawaban
catatan grafik irisan:
|
![{\displaystyle 10-4{\sqrt {3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d9a86526fbd2720facc2083556dfee83753bed7) |
|
![{\displaystyle 10+4{\sqrt {3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6419cfc35f20a6998014666a245168ba0c15940) |
|
—— |
|
+++ |
|
——
|
- grafik irisan arsiran 1, 2 dan 3
|
-3 |
|
![{\displaystyle 10-4{\sqrt {3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d9a86526fbd2720facc2083556dfee83753bed7) |
|
8 |
|
![{\displaystyle 10+4{\sqrt {3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6419cfc35f20a6998014666a245168ba0c15940) |
|
|
|
|
|
A |
|
A |
|
|
A |
|
A |
|
A |
|
|
|
|
|
|
A |
|
A |
|
A |
|
A
|
|
Vertikal |
Horisontal
|
|
Titik pusat (0,0)
|
Persamaan |
![{\displaystyle x^{2}=4py}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa3f7178d35b71078913a6dda45aa35c18b0d7d0) |
|
Sumbu simetri |
sumbu y |
sumbu x
|
Fokus |
![{\displaystyle F(0,p)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51f1f48e74076ff928a4c92030ed41944201e5f2) |
|
Direktris |
![{\displaystyle y=-p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56e9a2a0c0d50deaa16e19ff2ec86f3546c6f61d) |
|
|
Titik pusat (h,k)
|
Persamaan |
![{\displaystyle (x-h)^{2}=4p(y-k)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc6557f42b456c85644e99bdcb2fec5feec410e1) |
|
Sumbu simetri |
![{\displaystyle x=h}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99d77df94024e4133b5b9cd4b387cfb6548d198d) |
|
Fokus |
![{\displaystyle F(h,k+p)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/daa43bdd7ac9e3a94e49bbdbfb12bb49d28028cd) |
|
Direktris |
![{\displaystyle y=k-p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a86ace07bc705edd831f96c5ffef845f914f8cf) |
|
- bergradien
(
)
Vertikal |
Horisontal
|
Titik pusat (0,0)
|
![{\displaystyle y=mx-pm}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb8ea5ba14098e94d0b2eb2ec79fb0400f368540) |
|
Titik pusat (h,k)
|
![{\displaystyle (y-k)=m(x-h)-pm}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccf94672d5ad556452385db5b7f0544fa23035ab) |
|
- jika persamaan garis lurus bergradien sejajar maka
![{\displaystyle m_{2}=m_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a421d100f7813bdaa32bca9d049c3971985db028)
- jika persamaan garis lurus bergradien tegak lurus maka
![{\displaystyle m_{2}={\frac {-1}{m_{1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2a543f99096dae7356c08090282fdfa97fc2397)
- melalui titik
![{\displaystyle (x_{1},y_{1})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fc74086e56542bd28b46a84faaee3cebdd4a899)
dengan cara bagi adil
Vertikal |
Horisontal
|
Titik pusat (0,0)
|
![{\displaystyle xx_{1}=2py+2py_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b398456ae0933d79799f8c7c2d420f2e1b35d57) |
|
Titik pusat (h,k)
|
![{\displaystyle (x-h)(x_{1}-h)=2p(y-k)+2p(y_{1}-k)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17d692d8a9078168ff3ca07e0562650132fb4d7c) |
|
- jika titik
berada di dalam bentuknya maka ada 1 persamaan garis singgung (1 langkah).
- jika titik
berada di luar bentuknya maka ada 2 persamaan garis singgung (2 langkah).
contoh
- Titik pusat (0,0)
- Tentukan persamaan garis singgung yang bergradien 2 terhadap
!
jawab:
![{\displaystyle y^{2}=16x->y^{2}=4(4x){\text{ jadi }}p=4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8614af55fde381b9d0e8f11b8013cbaef6be44c9)
![{\displaystyle y=mx+{\frac {p}{m}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb9165d4b46166a40388fd2929f0574bef17090f)
![{\displaystyle y=2x+{\frac {4}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a57a50e046df716cff8031d1100fb1a42e8e00c9)
![{\displaystyle y=2x+2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83411135549c3e8f8c534c542ea317d4a6167820)
- Tentukan persamaan garis singgung yang melalui (4,8) terhadap
!
jawab:
![{\displaystyle y^{2}=16x->y^{2}=4(4x){\text{ jadi }}p=4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8614af55fde381b9d0e8f11b8013cbaef6be44c9)
(dalam)
dengan cara bagi adil
![{\displaystyle yy_{1}=2px+2px_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f1a095910b7fe5ddeb98acc98d1bc4aa09db6ec)
![{\displaystyle 8y=2(4)x+2(4)(4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29b1de386a32fe7bdeb8c3983779703bfc580f0c)
(dibagi 8)
![{\displaystyle y=x+4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c0942d12215c8d1816d3e5ab419a56863bee4bb)
- Tentukan persamaan garis singgung yang melalui (1,5) terhadap
!
jawab:
![{\displaystyle y^{2}=16x->y^{2}=4(4x){\text{ jadi }}p=4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8614af55fde381b9d0e8f11b8013cbaef6be44c9)
(luar)
dengan cara bagi adil
![{\displaystyle yy_{1}=2px+2px_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f1a095910b7fe5ddeb98acc98d1bc4aa09db6ec)
![{\displaystyle 5y=2(4)x+2(4)(1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bc9830aa992dd4dbdd21167c54c9f3a9dca63c5)
![{\displaystyle 5y=8x+8}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eddf2d4da0697aa3c16268ebf70d87a2879dd55c)
![{\displaystyle y={\frac {8}{5}}x+{\frac {8}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47fd8ec896c3d2ccb3f0c68acb153a85bfdecbbd)
masukkan lah
![{\displaystyle ({\frac {8}{5}}x+{\frac {8}{5}})^{2}=16x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f25d1e7f2dfb70f92a55db76bc0900816a4a90e6)
![{\displaystyle {\frac {64}{25}}x^{2}+{\frac {128}{25}}x+{\frac {64}{25}}-16x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a5a50b9dcbec984bacf56ca0db165e4cb60c873)
![{\displaystyle {\frac {64}{25}}x^{2}+{\frac {128}{25}}x+{\frac {64}{25}}-{\frac {400}{25}}x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf334a0f37689758ff6aa6bf5dc7d1e8e45f0279)
(dibagi 16/25)
![{\displaystyle 4x^{2}-17x+4=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8f885e8f44d5a2b1c5c34b2c4b70d1f85e1771b)
maka kita mencari nilai x
![{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00c22777378f9c594c71158fea8946f2495f2a28)
![{\displaystyle x={\frac {17\pm {\sqrt {289-256}}}{8}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6495a84b9671f0fd8c6f5db2eb4823b0c0b2747c)
![{\displaystyle x={\frac {17\pm {\sqrt {33}}}{8}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bfb8f59a8a68fa48c59e4f246d6aa268c9fe8f9)
atau ![{\displaystyle x_{2}={\frac {17-{\sqrt {33}}}{8}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d51f3cfc032e41d04919872756362ac1c6fb3e9)
maka kita mencari nilai y
- untuk
![{\displaystyle x_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8788bf85d532fa88d1fb25eff6ae382a601c308)
![{\displaystyle y_{1}={\frac {8}{5}}({\frac {17+{\sqrt {33}}}{8}})+{\frac {8}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0e92b71ec31bca031b0fc81f767cbfdf550b38e)
![{\displaystyle y_{1}={\frac {17}{5}}+{\frac {\sqrt {33}}{5}}+{\frac {8}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a2c7c3e4485742207e3a683df23594ad31ca85c)
![{\displaystyle y_{1}=5+{\frac {\sqrt {33}}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0499c9b0e0ef4e063fe06c9fb2d10aa008785c9c)
jadi
- untuk
![{\displaystyle x_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7af1b928f06e4c7e3e8ebfd60704656719bd766)
![{\displaystyle y_{2}={\frac {8}{5}}({\frac {17-{\sqrt {33}}}{8}})+{\frac {8}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7da41bb9c9bbbc22bcafe29f64ffd9beb692dd0b)
![{\displaystyle y_{2}={\frac {17}{5}}-{\frac {\sqrt {33}}{5}}+{\frac {8}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d27d64e6e4f69b1c40816b39db250bd017f8ffb7)
![{\displaystyle y_{2}=5-{\frac {\sqrt {33}}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30bee0156847ec7958c704a18bd2cd260d64eacf)
jadi
kembali dengan cara bagi adil
- untuk persamaan singgung pertama
![{\displaystyle yy_{1}=2px+2px_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f1a095910b7fe5ddeb98acc98d1bc4aa09db6ec)
![{\displaystyle (5+{\frac {\sqrt {33}}{5}})y=2(4)x+2(4)({\frac {17+{\sqrt {33}}}{8}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c89b1ee5e214f4b5dda22fda36fb6f22c575f8ce)
![{\displaystyle (5+{\frac {\sqrt {33}}{5}})y=8x+17+{\sqrt {33}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f751b37565b38cc507992b5ad9d5e948861d5cfe)
- untuk persamaan singgung kedua
![{\displaystyle yy_{2}=2px+2px_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69020d92649c568586fc8b2b94f3896478ccd4ef)
![{\displaystyle (5-{\frac {\sqrt {33}}{5}})y=2(4)x+2(4)({\frac {17-{\sqrt {33}}}{8}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/780aade712d080c3dab272b34fb400f3d36e0fff)
![{\displaystyle (5-{\frac {\sqrt {33}}{5}})y=8x+17-{\sqrt {33}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e04a05aff41bcf9d41a5a13ead64be886d71b3b8)
- Titik pusat (h,k)
- Tentukan persamaan garis singgung
melalui persamaan yang tegak lurus
!
jawab:
ubah ke bentuk sederhana
![{\displaystyle y^{2}-6y-8x+9=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e2aa603dea12dddc98204dcecf8ec862b5d2d2b)
![{\displaystyle y^{2}-6y+9=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ee0e75cb0abf8ce78674ce26d29dd94465b45e3)
![{\displaystyle (y-3)^{2}=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/240df7e655e1e9a1d926e2b24d89a00e49c3cc17)
cari gradien persamaan
![{\displaystyle y-2x-5=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2bd241a112277c6567ae8c2b41122b11b9322da)
![{\displaystyle y=2x+5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0dcfd6c110b33eb69951410bda4e5576e56b2000)
gradien (
) = 2 karena tegak lurus menjadi
cari
![{\displaystyle (y-3)^{2}=8x->(y-3)^{2}=4(2x){\text{ jadi }}p=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7b5f56632061d29a4e48c7693626e1333662382)
![{\displaystyle y=mx+{\frac {p}{m}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb9165d4b46166a40388fd2929f0574bef17090f)
![{\displaystyle y=-{\frac {1}{2}}x+{\frac {2}{-{\frac {1}{2}}}}->y=-{\frac {1}{2}}x-4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6eac42016ff58db1ffae7c6e4ecc92d39c136d3)
- Tentukan persamaan garis singgung
yang berordinat 6!
jawab:
ubah ke bentuk sederhana
![{\displaystyle y^{2}-6y-8x+9=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e2aa603dea12dddc98204dcecf8ec862b5d2d2b)
![{\displaystyle y^{2}-6y+9=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ee0e75cb0abf8ce78674ce26d29dd94465b45e3)
![{\displaystyle (y-3)^{2}=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/240df7e655e1e9a1d926e2b24d89a00e49c3cc17)
cari absis dimana ordinat 6
![{\displaystyle (y-3)^{2}=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/240df7e655e1e9a1d926e2b24d89a00e49c3cc17)
![{\displaystyle (6-3)^{2}=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a358bbe7b2e8919ad28d681ec5e1a58a49b346aa)
![{\displaystyle 9=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c98740c82d9e30d73bbe4063b8de96c8f1b339ff)
![{\displaystyle x={\frac {9}{8}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b51db05bfab94ff42e8c59d033e0376f765e0e12)
![{\displaystyle (y-3)^{2}=8x->(y-3)^{2}=4(2x){\text{ jadi }}p=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7b5f56632061d29a4e48c7693626e1333662382)
dengan cara bagi adil
![{\displaystyle (y-k)(y_{1}-k)=2px+2px_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f8ca0c8558ac9c175e30ffbda787008499cd172)
![{\displaystyle (y-3)(6-3)=2(2)x+2(2)({\frac {9}{8}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ada69613fc6d75deaa4b91e6fc089fbb8f65819)
![{\displaystyle (y-3)3=4x+{\frac {9}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99870e3cb79712f770be5af8d9eb11e645be3298)
![{\displaystyle 3y-9=4x+{\frac {9}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f1190603aa5a76c123d7fe2a454991c7139e2bf)
![{\displaystyle 3y=4x+{\frac {27}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/530e005bd88cc5440d7617d0b12901d4d0309381)
![{\displaystyle y={\frac {4}{3}}x+{\frac {27}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adabfc49ee9ef63d51dbec5136b036f79e1764e6)
- Tentukan persamaan garis singgung yang melalui (1,6) terhadap
!
ubah ke bentuk sederhana
![{\displaystyle y^{2}-6y-8x+9=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e2aa603dea12dddc98204dcecf8ec862b5d2d2b)
![{\displaystyle y^{2}-6y+9=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ee0e75cb0abf8ce78674ce26d29dd94465b45e3)
![{\displaystyle (y-3)^{2}=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/240df7e655e1e9a1d926e2b24d89a00e49c3cc17)
![{\displaystyle (y-3)^{2}=8x->(y-3)^{2}=4(2x){\text{ jadi }}p=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7b5f56632061d29a4e48c7693626e1333662382)
(luar)
dengan cara bagi adil
![{\displaystyle (y-k)(y_{1}-k)=2px+2px_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f8ca0c8558ac9c175e30ffbda787008499cd172)
![{\displaystyle (y-3)(6-3)=2(2)x+2(2)(1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2346d7db4bacddd929a90195b597ee6547fab436)
![{\displaystyle (y-3)3=4x+4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c8f623fdecb6b0db638e615545d394d836d1477)
![{\displaystyle 3y-9=4x+4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66108e6df3b0c2d2045f6d9035c152a9f2a94a4e)
![{\displaystyle 3y=4x+13}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad24a31147c9c69495988accfbd103715e1f0eb2)
![{\displaystyle y={\frac {4}{3}}x+{\frac {13}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12dbb8ba28c2bfc1ae58691fc6a403124e728de6)
masukkan lah
![{\displaystyle ({\frac {4}{3}}x+{\frac {13}{3}}-3)^{2}=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1af6b788a9eedcff75a0042138135c832d41301)
![{\displaystyle ({\frac {4}{3}}x+{\frac {4}{3}})^{2}=8x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f92a7d1b3fa47d1053e54bfa2cda3a4d911f2214)
![{\displaystyle {\frac {16}{9}}x^{2}+{\frac {32}{9}}x+{\frac {16}{9}}-8x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/336688a17fb8cbe08af6aac44c2f54d97bdf8739)
(dibagi 8/9)
![{\displaystyle 2x^{2}+5x+2=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc5b99cf574f108b175eb91974a5b91be39cd8fa)
maka kita mencari nilai x
![{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00c22777378f9c594c71158fea8946f2495f2a28)
![{\displaystyle x={\frac {5\pm {\sqrt {25-16}}}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8321ba86c96fa576137931183ec7986900c24ecb)
![{\displaystyle x={\frac {5\pm {\sqrt {9}}}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b028dca1d076bfafafb49b91b79274144b444863)
atau ![{\displaystyle x_{2}={\frac {5-{\sqrt {9}}}{4}}={\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02ea9596400132d833a0fc0457bbec37a8b0a5c8)
maka kita mencari nilai y
- untuk
![{\displaystyle x_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8788bf85d532fa88d1fb25eff6ae382a601c308)
![{\displaystyle y_{1}={\frac {4}{3}}(2)+{\frac {13}{3}}={\frac {8}{3}}+{\frac {13}{3}}=7}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e42d1433b26ca18913e5ad4164720bcbf610c55)
jadi
- untuk
![{\displaystyle x_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7af1b928f06e4c7e3e8ebfd60704656719bd766)
![{\displaystyle y_{2}={\frac {4}{3}}({\frac {1}{2}})+{\frac {13}{3}}={\frac {2}{3}}+{\frac {13}{3}}=5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aefd6621d25e8c5a514cde0de054dd44f35c05f1)
jadi
kembali dengan cara bagi adil
- untuk persamaan singgung pertama
![{\displaystyle (y-k)(y_{1}-k)=2px+2px_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f8ca0c8558ac9c175e30ffbda787008499cd172)
![{\displaystyle (y-3)(7-3)=2(2)x+2(2)(2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f025f3a07be29cc3c6174d91970d2402d14c20c)
![{\displaystyle (y-3)4=4x+8}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ea8ec966d063abcb9f237a2271e95281d7c7d53)
![{\displaystyle 4y-12=4x+8}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf222f900b2aa1f82edb9745e0b168a7a118f44d)
(dibagi 4)
![{\displaystyle y=x+5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bd098373ac90877f4dd1fb6753ffbe1728ce73b)
- untuk persamaan singgung kedua
![{\displaystyle (y-k)(y_{2}-k)=2px+2px_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9444bbf740f46b996ecc2c2fd51e84f8a7e6251f)
![{\displaystyle (y-3)(5-3)=2(2)x+2(2)({\frac {1}{2}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a0a43798b4d5c2654c313a29c1e1ed00a645602)
![{\displaystyle (y-3)2=4x+2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04bc94d92db132acaf9c997635375bd39594fa46)
![{\displaystyle 2y-6=4x+2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bee58f3e5444ca331b04a448387ad162b3c0fe4b)
(dibagi 2)
![{\displaystyle y=2x+4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8ce95b8668dabacc3dfa077b4e83350da3357cf)