x^2+6x+9=12 eching

$$x ^ { 2 } + 6 x + 9 = 12$$

x uchun yechish

$$x=2\sqrt{3}-3\approx 0,464101615$$

$$x=-2\sqrt{3}-3\approx -6,464101615$$

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.

$$x^{2}+6x+9=12$$

Tenglamaning ikkala tarafidan 12 ni ayirish.

$$x^{2}+6x+9-12=12-12$$

O‘zidan 12 ayirilsa 0 qoladi.

$$x^{2}+6x+9-12=0$$

9 dan 12 ni ayirish.

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

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

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

6 kvadratini chiqarish.

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

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+12}}{2}$$

36 ni 12 ga qo'shish.

$$x=\frac{-6±\sqrt{48}}{2}$$

48 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±4\sqrt{3}}{2}$$

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

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

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

$$x=2\sqrt{3}-3$$

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

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

-6-4\sqrt{3} ni 2 ga bo'lish.

$$x=-2\sqrt{3}-3$$

Tenglama yechildi.

$$x=2\sqrt{3}-3 x=-2\sqrt{3}-3$$

Kvadratni yakunlash bo'yicha qadamlar

x^{2}+6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{12}$$

Qisqartirish.

$$x+3=2\sqrt{3} x+3=-2\sqrt{3}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=2\sqrt{3}-3 x=-2\sqrt{3}-3$$