x^2+2x=-3 eching

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

x uchun yechish (complex solution)

$$x=-1+\sqrt{2}i\approx -1+1,414213562i$$

$$x=-\sqrt{2}i-1\approx -1-1,414213562i$$

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.

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

3 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -3 ayirilsa 0 qoladi.

$$x^{2}+2x-\left(-3\right)=0$$

0 dan -3 ni ayirish.

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

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

$$x=\frac{-2±\sqrt{2^{2}-4\times 3}}{2}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\times 3}}{2}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-12}}{2}$$

4 ni -12 ga qo'shish.

$$x=\frac{-2±\sqrt{-8}}{2}$$

-8 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{2}i}{2}$$

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

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

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

$$x=-1+\sqrt{2}i$$

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

$$x=\frac{-2\sqrt{2}i-2}{2}$$

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

$$x=-\sqrt{2}i-1$$

Tenglama yechildi.

$$x=-1+\sqrt{2}i x=-\sqrt{2}i-1$$

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.

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

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.

$$x^{2}+2x+1^{2}=-3+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=-3+1$$

-3 ni 1 ga qo'shish.

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

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+1=\sqrt{2}i x+1=-\sqrt{2}i$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=-1+\sqrt{2}i x=-\sqrt{2}i-1$$