Persamaan dan Fungsi Kuadrat

A.Persamaan Kuadrat

Persamaan kuadrat adalah kalimat terbuka
dengan pangkat tertinggi variabelnya 2.
Bentuk umum persamaan kuadrat:

${ax}^{2}+bx+c=0,a,b,c \epsilon R$ dan $a\neq 0$

Contoh:

${x}^{2}+3x+2=0$ adalah persamaan kuadrat dengan $a=1,c=3,c=2$

$-3{x}^{2}+7x-2=0$ adalah persamaan kuadrat dengan $a=-3,b=7,c=-2$

Cara menyelesaikan persamaan kuadrat (mencari akar-akar persamaan kuadrat):

1.Faktorisasi

${x}^{2} +2xy={y}^{2}=0$

$\left(x+y \right){}^{2}=0$

Contoh:

${x}^{2}+6x+9=0$

${x}^{2}+2.3x+{3}^{2}=0$

$\left(x+3 \right){}^{2}=0$

$x+2=0$

${x}_{1,2}=-3$

$2f+16t+32=0$

$2\left(f+8t+16 \right)=0$

$2\left(f+2.4t+{4}^{2} \right)=0$

$2\left(t+4 \right){}^{2}=0$

$t+2=0$

${t}_{1,2}=-4$

${x}^{2}-2xy+{y}^{2}=0$

$\left(x-y \right){}^{2}=0$

Contoh:

${x}^{2}-10x+25=0$

${x}^{2}-2.5x+\left(-5 \right){}^{2}=0$

$\left(x-5 \right){}^{2}=0$

$x-5=0$

${x}_{1,2}=5$

${-3s}^{2}+18s-27=0$

$-3\left({s}^{2}-6s+9 \right)=0$

$-3\left({s}^{2}-2.3s+\left(-3 \right){}^{2} \right)=0$

$-3\left(s-3 \right){}^{2}=0$

$s-3=0$

${s}_{1,2}=3$

${x}^{2}-{y}^{2}=0$

$\left(x+y \right)\left(x-y \right)=0$

Contoh:

${s}^{2}-25=0$

${s}^{2}-{5}^{2}=0$

$\left(s+5 \right)\left(s-5 \right)=0$

$s+5=0.atau.s-5=0$

${s}_{1}=-5........{s}_{2}=5$

${3r}^{2}-108=0$

$3\left({r}^{2}-36\right)=0$

$\left({r}^{2}-{6}^{2} \right)=0$

$\left(r+6 \right)\left(r-6 \right)=0$

$r+6=0.atau.r-6=0$

${r}_{1}=-6.........{r}_{2}=6$

${ax}^{2}+bx+c=0.dengan.a=1$

${ax}^{2}+bx+c=\left(x+p \right)\left(x+q \right)=0$

$dengan syarat: c=pq$

$b=p+q$

Contoh:

${x}^{2}+5x+6=0$

$\left(x+2 \right)\left(x+3 \right)=0$

$x+2=0.atau.x+3=0$

${x}_{1}=-2..........{x}_{2}=3$

${x}^{2}+8x-9=0$

$\left(x-1 \right)\left(x+9 \right)=0$

$x-1.atau.x+9=0$

${x}_{1}=1........{x}_{2}=-9$

${ax}^{2}+bx+c=0.dengan, a\neq 1$

${ax}^{2}+bx+c={ax}^{2}+px+qx+c=0$

dengan syarat:

$ac=pq$

$b=p+q$

Contoh:

${2x}^{2}+7x+5=0$

${2x}^{2}2x+5x+5=0$

$2x\left(x+1 \right)+5\left(x+1 \right)=0$

$\left(2x+5 \right)\left(x+1 \right)=0$

$2x-5=0.atau.x+1=0$

${x}_{1}=-\frac{5}{2}...................{c}_{2}=-1$

${-3x}^{2}+10x-7=0$

${-3x}^{2}+3x-7x-7=0$

$-3x\left(x-1 \right)+7\left(x-1 \right)=0$

$\left(-3x+7 \right)\left(x-1 \right)=0$

${x}_{1}=\frac{7}{3}..........{x}_{2}=1$

2.Kuadrat Sempurna

${x}^{2}+2bx=c$

${x}^{2}+2bx+{b}^{2}=c+{b}^{2}$

$\left(x+b \right){}^{2}=c+{b}^{2}$

$x=-b+_ \sqrt{c+{b}^{2}}$

Contoh:

${x}^{2}+6x=1$

${x}^{2}+6x+9=1+9$

$\left(x+3 \right){}^{2}=10$

$x+3=+-\sqrt{10}$

${x}_{1,2}=-3+-\sqrt{10}$

${x}_{1}=-3+\sqrt{10}.atau,{x}_{2}=-3-\sqrt{10}$

${x}^{2}-8x=-7$

${x}^{2}-8x+16=-7+16$

$\left(x-4 \right){}^{2}=9$

$x-4=+-\sqrt{9}$

${x}_{1,2}=4+-3$

${x}_{1}=4+3=7.atau.{x}_{2}=4-3=1$

3.Rumus abc

${ax}^{2}+bx+c=0$

${x}_{1,2}=\frac{-b+-\sqrt{{b}^{2}-4ac}}{2a}$

Contoh:

${3x}^{2}+7x+4=0$

${x}_{1,2}=\frac{-b+-\sqrt{{b}^{2}-4ac}}{2a}$

${x}_{1,2}=\frac{-7+-\sqrt{{7}^{2}-4.\left(3 \right).\left(4 \right)}}{2\left(3 \right)}$

$=\frac{-7+-\sqrt{1}}{6}$

${x}_{1}=\frac{-7+1}{6}=-1$

${x}_{2}=\frac{-7-1}{6}=-\frac{8}{6}$

${-2x}^{2}+x+2=0$

${x}_{1,2}=\frac{-b+-\sqrt{{b}^{2}-4ac}}{2a}$

${x}_{1,2}=\frac{-1+-\sqrt{{1}^{2}-4.\left(2 \right).\left(2 \right)}}{2\left(-2 \right)}$

$=\frac{1+-\sqrt{17}}{-4}$

${x}_{1}=\frac{-1-\sqrt{17}}{4}$

${x}_{2}=\frac{1+\sqrt{17}}{4}$

Berdasarkan rumus abc, ${ax}^{2}+bx+c=0,a,b,c\epsilon,dan,a\neq 0$ memiliki akar-akar

${x}_{1,2}=\frac{-b+-\sqrt{D}}{2a}dengan,D={b}^{2}-4ac.$

Jenis-jenis akar persamaan kuadrat:

-Jika D=0 atau ${b}^{2}-4ac$ ,memiliki 2 akar real yang sama/kembar ${x}_{1}={x}_{2}=\frac{-b}{2a}$

-Jika D>0,2 akar real beda ${x}_{1}\frac{-b-\sqrt{D}}{2a}dan,{x}_{2}=\frac{-b+\sqrt{D}}{2a}$

-Jika D<0,akar-akarnya bilangan kompleks
- $b=0,2.akar berlawanan.{x}_{1}={-x}_{2}$

- $a=c,2.akar berkebalikan.{x}_{1}=\frac{1}{{x}_{2}}$

Contoh:

${x}^{2}+3x+1=0$

Buktikan pesamaan kuadrat yang memiliki 2 akar berkebalikan ${x}_{1}=\frac{1}{{x}_{2}}$

${x}_{1,2}=\frac{-3+-\sqrt{{3}^{2}-4\left(1 \right)\left(1 \right)}}{2\left(1 \right)}$

${x}_{1,2}=\frac{-3+-\sqrt{5}}{2}$

${x}_{1}=\frac{-3+\sqrt{5}}{2}$

${x}_{2}=\frac{-3-\sqrt{5}}{2}$

${x}_{1}{x}_{2}=\left(\frac{-3+\sqrt{5}}{2} \right)\left(\frac{-3-\sqrt{5}}{2} \right)=\frac{9-5}{4}=1$

Karena ${x}_{1}{x}_{2}=1$ ,terbukti ${x}_{1}$ dan ${x}_{2}$ berkebalikan.

rumus2 persamaan pangkat 2 ada yg bisa melengkapi?

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