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

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

x uchun yechish

$$x=\frac{\sqrt{15}}{3}-1\approx 0,290994449$$

$$x=-\frac{\sqrt{15}}{3}-1\approx -2,290994449$$

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.

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

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

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

6 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-12\left(-2\right)}}{2\times 3}$$

-12 ni -2 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+24}}{2\times 3}$$

36 ni 24 ga qo'shish.

$$x=\frac{-6±\sqrt{60}}{2\times 3}$$

60 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{15}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{15}}{6}$$

x=\frac{-6±2\sqrt{15}}{6} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{15} ga qo'shish.

$$x=\frac{2\sqrt{15}-6}{6}$$

-6+2\sqrt{15} ni 6 ga bo'lish.

$$x=\frac{\sqrt{15}}{3}-1$$

x=\frac{-6±2\sqrt{15}}{6} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{15} ni ayirish.

$$x=\frac{-2\sqrt{15}-6}{6}$$

-6-2\sqrt{15} ni 6 ga bo'lish.

$$x=-\frac{\sqrt{15}}{3}-1$$

Tenglama yechildi.

$$x=\frac{\sqrt{15}}{3}-1 x=-\frac{\sqrt{15}}{3}-1$$

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.

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

2 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -2 ayirilsa 0 qoladi.

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

0 dan -2 ni ayirish.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

6 ni 3 ga bo'lish.

$$x^{2}+2x=\frac{2}{3}$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

1 kvadratini chiqarish.

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

\frac{2}{3} ni 1 ga qo'shish.

$$x^{2}+2x+1=\frac{5}{3}$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+1\right)^{2}=\frac{5}{3}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{5}{3}}$$

Qisqartirish.

$$x+1=\frac{\sqrt{15}}{3} x+1=-\frac{\sqrt{15}}{3}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\frac{\sqrt{15}}{3}-1 x=-\frac{\sqrt{15}}{3}-1$$