2z^2-2z+5=0 eching

$$2 z ^ { 2 } - 2 z + 5 = 0$$

z uchun yechish

$$z=\frac{1}{2}+\frac{3}{2}i=0,5+1,5i$$

$$z=\frac{1}{2}-\frac{3}{2}i=0,5-1,5i$$

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.

$$2z^{2}-2z+5=0$$

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

$$z=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\times 5}}{2\times 2}$$

-2 kvadratini chiqarish.

$$z=\frac{-\left(-2\right)±\sqrt{4-4\times 2\times 5}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$z=\frac{-\left(-2\right)±\sqrt{4-8\times 5}}{2\times 2}$$

-8 ni 5 marotabaga ko'paytirish.

$$z=\frac{-\left(-2\right)±\sqrt{4-40}}{2\times 2}$$

4 ni -40 ga qo'shish.

$$z=\frac{-\left(-2\right)±\sqrt{-36}}{2\times 2}$$

-36 ning kvadrat ildizini chiqarish.

$$z=\frac{-\left(-2\right)±6i}{2\times 2}$$

-2 ning teskarisi 2 ga teng.

$$z=\frac{2±6i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$z=\frac{2±6i}{4}$$

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

$$z=\frac{2+6i}{4}$$

2+6i ni 4 ga bo'lish.

$$z=\frac{1}{2}+\frac{3}{2}i$$

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

$$z=\frac{2-6i}{4}$$

2-6i ni 4 ga bo'lish.

$$z=\frac{1}{2}-\frac{3}{2}i$$

Tenglama yechildi.

$$z=\frac{1}{2}+\frac{3}{2}i z=\frac{1}{2}-\frac{3}{2}i$$

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.

$$2z^{2}-2z+5=0$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$2z^{2}-2z+5-5=-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$2z^{2}-2z=-5$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2z^{2}-2z}{2}=-\frac{5}{2}$$

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

$$z^{2}+\left(-\frac{2}{2}\right)z=-\frac{5}{2}$$

-2 ni 2 ga bo'lish.

$$z^{2}-z=-\frac{5}{2}$$

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

$$z^{2}-z+\left(-\frac{1}{2}\right)^{2}=-\frac{5}{2}+\left(-\frac{1}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

$$z^{2}-z+\frac{1}{4}=-\frac{5}{2}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{5}{2} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$z^{2}-z+\frac{1}{4}=-\frac{9}{4}$$

z^{2}-z+\frac{1}{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.

$$\left(z-\frac{1}{2}\right)^{2}=-\frac{9}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(z-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{9}{4}}$$

Qisqartirish.

$$z-\frac{1}{2}=\frac{3}{2}i z-\frac{1}{2}=-\frac{3}{2}i$$

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$z=\frac{1}{2}+\frac{3}{2}i z=\frac{1}{2}-\frac{3}{2}i$$