3x^2+6x=12 eching

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

x uchun yechish (complex solution)

$$x=\sqrt{5}-1\approx 1,236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3,236067977$$

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.

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

Tenglamaning ikkala tarafidan 12 ni ayirish.

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

O‘zidan 12 ayirilsa 0 qoladi.

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

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 -12 ni c bilan almashtiring.

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

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 3\left(-12\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

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

-12 ni -12 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+144}}{2\times 3}$$

36 ni 144 ga qo'shish.

$$x=\frac{-6±\sqrt{180}}{2\times 3}$$

180 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±6\sqrt{5}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-6±6\sqrt{5}}{6}$$

x=\frac{-6±6\sqrt{5}}{6} tenglamasini yeching, bunda ± musbat. -6 ni 6\sqrt{5} ga qo'shish.

$$x=\frac{6\sqrt{5}-6}{6}$$

-6+6\sqrt{5} ni 6 ga bo'lish.

$$x=\sqrt{5}-1$$

x=\frac{-6±6\sqrt{5}}{6} tenglamasini yeching, bunda ± manfiy. -6 dan 6\sqrt{5} ni ayirish.

$$x=\frac{-6\sqrt{5}-6}{6}$$

-6-6\sqrt{5} ni 6 ga bo'lish.

$$x=-\sqrt{5}-1$$

Tenglama yechildi.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

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.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

6 ni 3 ga bo'lish.

$$x^{2}+2x=\frac{12}{3}$$

12 ni 3 ga bo'lish.

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

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=4+1$$

4 ni 1 ga qo'shish.

$$x^{2}+2x+1=5$$

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

x uchun yechish

$$x=\sqrt{5}-1\approx 1,236067977$$

$$x=-\sqrt{5}-1\approx -3,236067977$$

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.

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

Tenglamaning ikkala tarafidan 12 ni ayirish.

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

O‘zidan 12 ayirilsa 0 qoladi.

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

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 -12 ni c bilan almashtiring.

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

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 3\left(-12\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

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

-12 ni -12 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+144}}{2\times 3}$$

36 ni 144 ga qo'shish.

$$x=\frac{-6±\sqrt{180}}{2\times 3}$$

180 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±6\sqrt{5}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-6±6\sqrt{5}}{6}$$

x=\frac{-6±6\sqrt{5}}{6} tenglamasini yeching, bunda ± musbat. -6 ni 6\sqrt{5} ga qo'shish.

$$x=\frac{6\sqrt{5}-6}{6}$$

-6+6\sqrt{5} ni 6 ga bo'lish.

$$x=\sqrt{5}-1$$

x=\frac{-6±6\sqrt{5}}{6} tenglamasini yeching, bunda ± manfiy. -6 dan 6\sqrt{5} ni ayirish.

$$x=\frac{-6\sqrt{5}-6}{6}$$

-6-6\sqrt{5} ni 6 ga bo'lish.

$$x=-\sqrt{5}-1$$

Tenglama yechildi.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

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.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

6 ni 3 ga bo'lish.

$$x^{2}+2x=\frac{12}{3}$$

12 ni 3 ga bo'lish.

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

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=4+1$$

4 ni 1 ga qo'shish.

$$x^{2}+2x+1=5$$

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$