3x^2-12x+6=0 eching

$$3 x ^ { 2 } - 12 x + 6 = 0$$

x uchun yechish

$$x=\sqrt{2}+2\approx 3,414213562$$

$$x=2-\sqrt{2}\approx 0,585786438$$

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.

$$3x^{2}-12x+6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -12 ni b va 6 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\times 6}}{2\times 3}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 3\times 6}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-12\times 6}}{2\times 3}$$

-12 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-72}}{2\times 3}$$

144 ni -72 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{72}}{2\times 3}$$

72 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±6\sqrt{2}}{2\times 3}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±6\sqrt{2}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{12±6\sqrt{2}}{6}$$

x=\frac{12±6\sqrt{2}}{6} tenglamasini yeching, bunda ± musbat. 12 ni 6\sqrt{2} ga qo'shish.

$$x=\frac{6\sqrt{2}+12}{6}$$

12+6\sqrt{2} ni 6 ga bo'lish.

$$x=\sqrt{2}+2$$

x=\frac{12±6\sqrt{2}}{6} tenglamasini yeching, bunda ± manfiy. 12 dan 6\sqrt{2} ni ayirish.

$$x=\frac{12-6\sqrt{2}}{6}$$

12-6\sqrt{2} ni 6 ga bo'lish.

$$x=2-\sqrt{2}$$

Tenglama yechildi.

$$x=\sqrt{2}+2 x=2-\sqrt{2}$$

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.

$$3x^{2}-12x+6=0$$

Tenglamaning ikkala tarafidan 6 ni ayirish.

$$3x^{2}-12x+6-6=-6$$

O‘zidan 6 ayirilsa 0 qoladi.

$$3x^{2}-12x=-6$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-12x}{3}=-\frac{6}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{12}{3}\right)x=-\frac{6}{3}$$

-12 ni 3 ga bo'lish.

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

-6 ni 3 ga bo'lish.

$$x^{2}-4x=-2$$

-4 ni bo‘lish, x shartining koeffitsienti, 2 ga -2 olish uchun. Keyin, -2 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

-2 kvadratini chiqarish.

$$x^{2}-4x+4=-2+4$$

-2 ni 4 ga qo'shish.

$$x^{2}-4x+4=2$$

x^{2}-4x+4 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-2=\sqrt{2} x-2=-\sqrt{2}$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{2}+2 x=2-\sqrt{2}$$