x^2+4x+68=0 eching

$$x ^ { 2 } + 4 x + 68 = 0$$

x uchun yechish (complex solution)

$$x=-2+8i$$

$$x=-2-8i$$

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.

$$x^{2}+4x+68=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 4 ni b va 68 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 68}}{2}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 68}}{2}$$

-4 ni 68 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-272}}{2}$$

16 ni -272 ga qo'shish.

$$x=\frac{-4±\sqrt{-256}}{2}$$

-256 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±16i}{2}$$

x=\frac{-4±16i}{2} tenglamasini yeching, bunda ± musbat. -4 ni 16i ga qo'shish.

$$x=\frac{-4+16i}{2}$$

-4+16i ni 2 ga bo'lish.

$$x=-2+8i$$

x=\frac{-4±16i}{2} tenglamasini yeching, bunda ± manfiy. -4 dan 16i ni ayirish.

$$x=\frac{-4-16i}{2}$$

-4-16i ni 2 ga bo'lish.

$$x=-2-8i$$

Tenglama yechildi.

$$x=-2+8i x=-2-8i$$

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.

$$x^{2}+4x+68=0$$

Tenglamaning ikkala tarafidan 68 ni ayirish.

$$x^{2}+4x+68-68=-68$$

O‘zidan 68 ayirilsa 0 qoladi.

$$x^{2}+4x=-68$$

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.

$$x^{2}+4x+2^{2}=-68+2^{2}$$

2 kvadratini chiqarish.

$$x^{2}+4x+4=-68+4$$

-68 ni 4 ga qo'shish.

$$x^{2}+4x+4=-64$$

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.

$$\left(x+2\right)^{2}=-64$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-64}$$

Qisqartirish.

$$x+2=8i x+2=-8i$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$x=-2+8i x=-2-8i$$