2x^2+x+3=0 eching

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

x uchun yechish (complex solution)

$$x=\frac{-1+\sqrt{23}i}{4}\approx -0,25+1,198957881i$$

$$x=\frac{-\sqrt{23}i-1}{4}\approx -0,25-1,198957881i$$

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.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\times 3}}{2\times 2}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 2\times 3}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-8\times 3}}{2\times 2}$$

-8 ni 3 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-24}}{2\times 2}$$

1 ni -24 ga qo'shish.

$$x=\frac{-1±\sqrt{-23}}{2\times 2}$$

-23 ning kvadrat ildizini chiqarish.

$$x=\frac{-1±\sqrt{23}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{23}i}{4}$$

x=\frac{-1±\sqrt{23}i}{4} tenglamasini yeching, bunda ± musbat. -1 ni i\sqrt{23} ga qo'shish.

$$x=\frac{-1+\sqrt{23}i}{4}$$

x=\frac{-1±\sqrt{23}i}{4} tenglamasini yeching, bunda ± manfiy. -1 dan i\sqrt{23} ni ayirish.

$$x=\frac{-\sqrt{23}i-1}{4}$$

Tenglama yechildi.

$$x=\frac{-1+\sqrt{23}i}{4} x=\frac{-\sqrt{23}i-1}{4}$$

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.

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

Tenglamaning ikkala tarafidan 3 ni ayirish.

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

O‘zidan 3 ayirilsa 0 qoladi.

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

Ikki tarafini 2 ga bo‘ling.

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

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

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

\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{4} olish uchun. Keyin, \frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{3}{2}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{3}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{23}{16}$$

x^{2}+\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{23}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{23}{16}}$$

Qisqartirish.

$$x+\frac{1}{4}=\frac{\sqrt{23}i}{4} x+\frac{1}{4}=-\frac{\sqrt{23}i}{4}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$x=\frac{-1+\sqrt{23}i}{4} x=\frac{-\sqrt{23}i-1}{4}$$