x^2+12x+64=0 eching

$${ x }^{ 2 } +12x+64=0$$

x uchun yechish (complex solution)

$$x=-6+2\sqrt{7}i\approx -6+5,291502622i$$

$$x=-2\sqrt{7}i-6\approx -6-5,291502622i$$

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.

$$x^{2}+12x+64=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 12 ni b va 64 ni c bilan almashtiring.

$$x=\frac{-12±\sqrt{12^{2}-4\times 64}}{2}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 64}}{2}$$

-4 ni 64 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-256}}{2}$$

144 ni -256 ga qo'shish.

$$x=\frac{-12±\sqrt{-112}}{2}$$

-112 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±4\sqrt{7}i}{2}$$

x=\frac{-12±4\sqrt{7}i}{2} tenglamasini yeching, bunda ± musbat. -12 ni 4i\sqrt{7} ga qo'shish.

$$x=\frac{-12+4\sqrt{7}i}{2}$$

-12+4i\sqrt{7} ni 2 ga bo'lish.

$$x=-6+2\sqrt{7}i$$

x=\frac{-12±4\sqrt{7}i}{2} tenglamasini yeching, bunda ± manfiy. -12 dan 4i\sqrt{7} ni ayirish.

$$x=\frac{-4\sqrt{7}i-12}{2}$$

-12-4i\sqrt{7} ni 2 ga bo'lish.

$$x=-2\sqrt{7}i-6$$

Tenglama yechildi.

$$x=-6+2\sqrt{7}i x=-2\sqrt{7}i-6$$

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.

$$x^{2}+12x+64=0$$

Tenglamaning ikkala tarafidan 64 ni ayirish.

$$x^{2}+12x+64-64=-64$$

O‘zidan 64 ayirilsa 0 qoladi.

$$x^{2}+12x=-64$$

12 ni bo‘lish, x shartining koeffitsienti, 2 ga 6 olish uchun. Keyin, 6 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+12x+6^{2}=-64+6^{2}$$

6 kvadratini chiqarish.

$$x^{2}+12x+36=-64+36$$

-64 ni 36 ga qo'shish.

$$x^{2}+12x+36=-28$$

x^{2}+12x+36 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+6\right)^{2}=-28$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+6\right)^{2}}=\sqrt{-28}$$

Qisqartirish.

$$x+6=2\sqrt{7}i x+6=-2\sqrt{7}i$$

Tenglamaning ikkala tarafidan 6 ni ayirish.

$$x=-6+2\sqrt{7}i x=-2\sqrt{7}i-6$$