x^2+6x=-11 eching

$$x ^ { 2 } + 6 x = - 11$$

x uchun yechish (complex solution)

$$x=-3+\sqrt{2}i\approx -3+1,414213562i$$

$$x=-\sqrt{2}i-3\approx -3-1,414213562i$$

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.

$$x^{2}+6x=-11$$

11 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+6x-\left(-11\right)=-11-\left(-11\right)$$

O‘zidan -11 ayirilsa 0 qoladi.

$$x^{2}+6x-\left(-11\right)=0$$

0 dan -11 ni ayirish.

$$x^{2}+6x+11=0$$

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 11 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 11}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 11}}{2}$$

-4 ni 11 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-44}}{2}$$

36 ni -44 ga qo'shish.

$$x=\frac{-6±\sqrt{-8}}{2}$$

-8 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{2}i}{2}$$

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

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

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

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

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

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

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

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

Tenglama yechildi.

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

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.

$$x^{2}+6x=-11$$

6 ni bo‘lish, x shartining koeffitsienti, 2 ga 3 olish uchun. Keyin, 3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+6x+3^{2}=-11+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=-11+9$$

-11 ni 9 ga qo'shish.

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

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan 3 ni ayirish.

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