1/3x^2+6x=9 eching

$$\frac { 1 } { 3 } x ^ { 2 } + 6 x = 9$$

x uchun yechish

$$x=6\sqrt{3}-9\approx 1,392304845$$

$$x=-6\sqrt{3}-9\approx -19,392304845$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$\frac{1}{3}x^{2}+6x=9$$

Tenglamaning ikkala tarafidan 9 ni ayirish.

$$\frac{1}{3}x^{2}+6x-9=9-9$$

O‘zidan 9 ayirilsa 0 qoladi.

$$\frac{1}{3}x^{2}+6x-9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} \frac{1}{3} ni a, 6 ni b va -9 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times \frac{1}{3}\left(-9\right)}}{2\times \frac{1}{3}}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times \frac{1}{3}\left(-9\right)}}{2\times \frac{1}{3}}$$

-4 ni \frac{1}{3} marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-\frac{4}{3}\left(-9\right)}}{2\times \frac{1}{3}}$$

-\frac{4}{3} ni -9 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+12}}{2\times \frac{1}{3}}$$

36 ni 12 ga qo'shish.

$$x=\frac{-6±\sqrt{48}}{2\times \frac{1}{3}}$$

48 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±4\sqrt{3}}{2\times \frac{1}{3}}$$

2 ni \frac{1}{3} marotabaga ko'paytirish.

$$x=\frac{-6±4\sqrt{3}}{\frac{2}{3}}$$

x=\frac{-6±4\sqrt{3}}{\frac{2}{3}} tenglamasini yeching, bunda ± musbat. -6 ni 4\sqrt{3} ga qo'shish.

$$x=\frac{4\sqrt{3}-6}{\frac{2}{3}}$$

-6+4\sqrt{3} ni \frac{2}{3} ga bo'lish -6+4\sqrt{3} ga k'paytirish \frac{2}{3} ga qaytarish.

$$x=6\sqrt{3}-9$$

x=\frac{-6±4\sqrt{3}}{\frac{2}{3}} tenglamasini yeching, bunda ± manfiy. -6 dan 4\sqrt{3} ni ayirish.

$$x=\frac{-4\sqrt{3}-6}{\frac{2}{3}}$$

-6-4\sqrt{3} ni \frac{2}{3} ga bo'lish -6-4\sqrt{3} ga k'paytirish \frac{2}{3} ga qaytarish.

$$x=-6\sqrt{3}-9$$

Tenglama yechildi.

$$x=6\sqrt{3}-9 x=-6\sqrt{3}-9$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$\frac{1}{3}x^{2}+6x=9$$

Ikkala tarafini 3 ga ko‘paytiring.

$$\frac{\frac{1}{3}x^{2}+6x}{\frac{1}{3}}=\frac{9}{\frac{1}{3}}$$

\frac{1}{3} ga bo'lish \frac{1}{3} ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{\frac{1}{3}}x=\frac{9}{\frac{1}{3}}$$

6 ni \frac{1}{3} ga bo'lish 6 ga k'paytirish \frac{1}{3} ga qaytarish.

$$x^{2}+18x=\frac{9}{\frac{1}{3}}$$

9 ni \frac{1}{3} ga bo'lish 9 ga k'paytirish \frac{1}{3} ga qaytarish.

$$x^{2}+18x=27$$

18 ni bo‘lish, x shartining koeffitsienti, 2 ga 9 olish uchun. Keyin, 9 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+18x+9^{2}=27+9^{2}$$

9 kvadratini chiqarish.

$$x^{2}+18x+81=27+81$$

27 ni 81 ga qo'shish.

$$x^{2}+18x+81=108$$

x^{2}+18x+81 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+9\right)^{2}=108$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+9\right)^{2}}=\sqrt{108}$$

Qisqartirish.

$$x+9=6\sqrt{3} x+9=-6\sqrt{3}$$

Tenglamaning ikkala tarafidan 9 ni ayirish.

$$x=6\sqrt{3}-9 x=-6\sqrt{3}-9$$