x=(x+2)^2-1 eching

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

x uchun yechish (complex solution)

$$x=\frac{-\sqrt{3}i-3}{2}\approx -1,5-0,866025404i$$

$$x=\frac{-3+\sqrt{3}i}{2}\approx -1,5+0,866025404i$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+2\right)^{2} kengaytirilishi uchun ishlating.

$$x=x^{2}+4x+4-1$$

3 olish uchun 4 dan 1 ni ayirish.

$$x=x^{2}+4x+3$$

Ikkala tarafdan x^{2} ni ayirish.

$$x-x^{2}=4x+3$$

Ikkala tarafdan 4x ni ayirish.

$$x-x^{2}-4x=3$$

-3x ni olish uchun x va -4x ni birlashtirish.

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

Ikkala tarafdan 3 ni ayirish.

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

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.

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

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

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\left(-3\right)}}{2\left(-1\right)}$$

4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-12}}{2\left(-1\right)}$$

9 ni -12 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{-3}}{2\left(-1\right)}$$

-3 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{3}i}{2\left(-1\right)}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±\sqrt{3}i}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{3±\sqrt{3}i}{-2}$$

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

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

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

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

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

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

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

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+2\right)^{2} kengaytirilishi uchun ishlating.

$$x=x^{2}+4x+4-1$$

3 olish uchun 4 dan 1 ni ayirish.

$$x=x^{2}+4x+3$$

Ikkala tarafdan x^{2} ni ayirish.

$$x-x^{2}=4x+3$$

Ikkala tarafdan 4x ni ayirish.

$$x-x^{2}-4x=3$$

-3x ni olish uchun x va -4x ni birlashtirish.

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

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

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

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}-3x}{-1}=\frac{3}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{3}{-1}\right)x=\frac{3}{-1}$$

-3 ni -1 ga bo'lish.

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

3 ni -1 ga bo'lish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{2} kvadratini chiqarish.

$$x^{2}+3x+\frac{9}{4}=-3+\frac{9}{4}$$

-3 ni \frac{9}{4} ga qo'shish.

$$x^{2}+3x+\frac{9}{4}=-\frac{3}{4}$$

x^{2}+3x+\frac{9}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

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