2x^2+4x+5=0 eching

$$2 { x }^{ 2 } +4x+5 = 0$$

x uchun yechish (complex solution)

$$x=\frac{\sqrt{6}i}{2}-1\approx -1+1,224744871i$$

$$x=-\frac{\sqrt{6}i}{2}-1\approx -1-1,224744871i$$

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.

$$2x^{2}+4x+5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 4 ni b va 5 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\times 5}}{2\times 2}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 2\times 5}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-8\times 5}}{2\times 2}$$

-8 ni 5 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-40}}{2\times 2}$$

16 ni -40 ga qo'shish.

$$x=\frac{-4±\sqrt{-24}}{2\times 2}$$

-24 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±2\sqrt{6}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±2\sqrt{6}i}{4}$$

x=\frac{-4±2\sqrt{6}i}{4} tenglamasini yeching, bunda ± musbat. -4 ni 2i\sqrt{6} ga qo'shish.

$$x=\frac{-4+2\sqrt{6}i}{4}$$

-4+2i\sqrt{6} ni 4 ga bo'lish.

$$x=\frac{\sqrt{6}i}{2}-1$$

x=\frac{-4±2\sqrt{6}i}{4} tenglamasini yeching, bunda ± manfiy. -4 dan 2i\sqrt{6} ni ayirish.

$$x=\frac{-2\sqrt{6}i-4}{4}$$

-4-2i\sqrt{6} ni 4 ga bo'lish.

$$x=-\frac{\sqrt{6}i}{2}-1$$

Tenglama yechildi.

$$x=\frac{\sqrt{6}i}{2}-1 x=-\frac{\sqrt{6}i}{2}-1$$

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.

$$2x^{2}+4x+5=0$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$2x^{2}+4x+5-5=-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$2x^{2}+4x=-5$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+4x}{2}=-\frac{5}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{4}{2}x=-\frac{5}{2}$$

4 ni 2 ga bo'lish.

$$x^{2}+2x=-\frac{5}{2}$$

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.

$$x^{2}+2x+1^{2}=-\frac{5}{2}+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=-\frac{5}{2}+1$$

-\frac{5}{2} ni 1 ga qo'shish.

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

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x+1=\frac{\sqrt{6}i}{2} x+1=-\frac{\sqrt{6}i}{2}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\frac{\sqrt{6}i}{2}-1 x=-\frac{\sqrt{6}i}{2}-1$$