2x^2+3x+2=0 eching

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

x uchun yechish (complex solution)

$$x=\frac{-3+\sqrt{7}i}{4}\approx -0,75+0,661437828i$$

$$x=\frac{-\sqrt{7}i-3}{4}\approx -0,75-0,661437828i$$

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.

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

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

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

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\times 2\times 2}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-8\times 2}}{2\times 2}$$

-8 ni 2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-16}}{2\times 2}$$

9 ni -16 ga qo'shish.

$$x=\frac{-3±\sqrt{-7}}{2\times 2}$$

-7 ning kvadrat ildizini chiqarish.

$$x=\frac{-3±\sqrt{7}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{7}i}{4}$$

x=\frac{-3±\sqrt{7}i}{4} tenglamasini yeching, bunda ± musbat. -3 ni i\sqrt{7} ga qo'shish.

$$x=\frac{-3+\sqrt{7}i}{4}$$

x=\frac{-3±\sqrt{7}i}{4} tenglamasini yeching, bunda ± manfiy. -3 dan i\sqrt{7} ni ayirish.

$$x=\frac{-\sqrt{7}i-3}{4}$$

Tenglama yechildi.

$$x=\frac{-3+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i-3}{4}$$

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.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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

O‘zidan 2 ayirilsa 0 qoladi.

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

Ikki tarafini 2 ga bo‘ling.

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

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

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

-2 ni 2 ga bo'lish.

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

\frac{3}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{4} olish uchun. Keyin, \frac{3}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{4} kvadratini chiqarish.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-1+\frac{9}{16}$$

-1 ni \frac{9}{16} ga qo'shish.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=-\frac{7}{16}$$

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

$$\left(x+\frac{3}{4}\right)^{2}=-\frac{7}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Qisqartirish.

$$x+\frac{3}{4}=\frac{\sqrt{7}i}{4} x+\frac{3}{4}=-\frac{\sqrt{7}i}{4}$$

Tenglamaning ikkala tarafidan \frac{3}{4} ni ayirish.

$$x=\frac{-3+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i-3}{4}$$