Question

Jika α dan β akar-akar persamaan 3x²+x+2=0 maka hitunglah soal diatas​

Answer 1

Jawaban:

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Penjelasan dengan langkah-langkah:

3x² + x + 2 = 0

a = 3

b = 1

c = 2

a. a² + b²

= (a + b)² - 2ab

= (-b/a)² - 2(c/a)

= (-1/3)² - 2(2/3)

= 1/9 - 4/3

= 1/9 - 12/9

= -11/9

b. (a - 2)² + (b - 2)²

= a² - 4a + 4 + b² - 4b + 4

= a² + b² - 4a - 4b + 8

= (a² + b²) - 4(a + b) + 8

= (a + b)² - 2ab - 4(a + b) + 8

= (-b/a)² - 2(c/a) - 4(-b/a) + 8

= (-1/3)² - 2(2/3) - 4(-1/3) + 8

= 1/9 - 4/3 + 4/3 + 8

= 1/9 + 8

= 73/9

a dan b dalam jawaban bermaksud alpha dan beta.

x1 + x2 = -b/a

x1 . x2 = c/a

Answer 2

Verified answer

Jika α dan β akar-akar persamaan 3x² + x + 2 = 0, maka:

[tex]\begin{aligned}\sf a.\ &\alpha^2+\beta^2=\,\boxed{\,\bf{-}\frac{11}{9}\,}\\\vphantom{\Bigg|}\sf b.\ &(\alpha-2)^2+(\beta-2)^2=\,\boxed{\,\bf\frac{73}{9}\,}\,=\,\boxed{\,\bf8\,\frac{1}{9}\,}\\\vphantom{\Bigg|}\sf c.\ &\alpha^2\beta+\alpha\beta^2=\,\boxed{\,\bf{-}\frac{2}{9}\,}\\\sf d.\ &\frac{1}{\alpha+1}+\frac{1}{\beta+1}=\,\boxed{\,\bf\frac{5}{4}\,}\,=\,\boxed{\,\bf1\,\frac{1}{4}\,}\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah:

Akar-akar Persamaan Kuadrat

Jika α dan β akar-akar persamaan 3x² + x + 2 = 0 maka:

• Jumlah akar-akarnya:
  α + β = –b/a = –1/3
• Hasil kali akar-akarnya:
  αβ = c/a = 2/3

Soal a

[tex]\begin{aligned}\alpha^2+\beta^2&=(\alpha+\beta)^2-2\alpha\beta\\&=\left(-\frac{1}{3}\right)^2-2\left(\frac{2}{3}\right)\\&=\frac{1}{9}-\frac{4}{3}\\&=\frac{1-12}{9}\\\alpha^2+\beta^2&=\boxed{\,\bf{-}\frac{11}{9}\,}\end{aligned}[/tex]

Soal b

[tex]\begin{aligned}&(\alpha-2)^2+(\beta-2)^2\\&{=\ }\alpha^2-4\alpha+4+\beta^2-4\beta+4\\&{=\ }\underbrace{\alpha^2+\beta^2}_{\begin{matrix}\sf soal\ a\end{matrix}}-4(\alpha+\beta)+8\\&{=\ }{-}\frac{11}{9}-4\left(-\frac{1}{3}\right)+8\\&{=\ }{-}\frac{11}{9}+\frac{4}{3}+8\\&{=\ }\frac{-11+12+72}{9}\\&{=\ }\frac{1+72}{9}\\&{=\ }\boxed{\,\bf\frac{73}{9}=8\,\frac{1}{9}\,}\end{aligned}[/tex]
_____________

Sekaligus untuk soal lanjutannya di pertanyaan lain.

Soal c

[tex]\begin{aligned}\alpha^2\beta+\alpha\beta^2&=(\alpha\beta)(\alpha+\beta)\\&=\frac{2}{3}\cdot\left(-\frac{1}{3}\right)\\\alpha^2\beta+\alpha\beta^2&=\boxed{\,\bf{-}\frac{2}{9}\,}\end{aligned}[/tex]

Soal d

[tex]\begin{aligned}&\frac{1}{\alpha+1}+\frac{1}{\beta+1}\\&{=\ }\frac{\beta+1}{(\alpha+1)(\beta+1)}+\frac{\alpha+1}{(\alpha+1)(\beta+1)}\\&{=\ }\frac{\alpha+1+\beta+1}{(\alpha+1)(\beta+1)}\\&{=\ }\frac{\alpha+\beta+2}{\alpha\beta+\alpha+\beta+1}\\&{=\ }\frac{-\dfrac{1}{3}+2}{\dfrac{2}{3}+\left(-\dfrac{1}{3}\right)+1}\\&{=\ }\frac{3\cdot\left(-\dfrac{1}{3}+2\right)}{3\cdot\left(\dfrac{2}{3}-\dfrac{1}{3}+1\right)}\\&{=\ }\frac{-1+6}{2-1+3}\\&{=\ }\boxed{\,\bf\frac{5}{4}=1\,\frac{1}{4}\,}\end{aligned}[/tex]
[tex]\blacksquare[/tex]