2x^2-3x+6=0 eching

$$2 x ^ { 2 } - 3 x + 6 = 0$$

x uchun yechish (complex solution)

$$x=\frac{3+\sqrt{39}i}{4}\approx 0,75+1,5612495i$$

$$x=\frac{-\sqrt{39}i+3}{4}\approx 0,75-1,5612495i$$

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.

$$2x^{2}-3x+6=0$$

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 6 ni c bilan almashtiring.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 6}}{2\times 2}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times 6}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\times 6}}{2\times 2}$$

-8 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-48}}{2\times 2}$$

9 ni -48 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{-39}}{2\times 2}$$

-39 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{39}i}{2\times 2}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±\sqrt{39}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{3±\sqrt{39}i}{4}$$

x=\frac{3±\sqrt{39}i}{4} tenglamasini yeching, bunda ± musbat. 3 ni i\sqrt{39} ga qo'shish.

$$x=\frac{3+\sqrt{39}i}{4}$$

x=\frac{3±\sqrt{39}i}{4} tenglamasini yeching, bunda ± manfiy. 3 dan i\sqrt{39} ni ayirish.

$$x=\frac{-\sqrt{39}i+3}{4}$$

Tenglama yechildi.

$$x=\frac{3+\sqrt{39}i}{4} x=\frac{-\sqrt{39}i+3}{4}$$

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.

$$2x^{2}-3x+6=0$$

Tenglamaning ikkala tarafidan 6 ni ayirish.

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

O‘zidan 6 ayirilsa 0 qoladi.

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

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-3x}{2}=-\frac{6}{2}$$

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

$$x^{2}-\frac{3}{2}x=-\frac{6}{2}$$

-6 ni 2 ga bo'lish.

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

-\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.

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

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

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

-3 ni \frac{9}{16} ga qo'shish.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{39}{16}$$

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.

$$\left(x-\frac{3}{4}\right)^{2}=-\frac{39}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{-\frac{39}{16}}$$

Qisqartirish.

$$x-\frac{3}{4}=\frac{\sqrt{39}i}{4} x-\frac{3}{4}=-\frac{\sqrt{39}i}{4}$$

\frac{3}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{3+\sqrt{39}i}{4} x=\frac{-\sqrt{39}i+3}{4}$$