3z^2+3z+20=0 eching

$$3 z ^ { 2 } + 3 z + 20 = 0$$

z uchun yechish

$$z=\frac{\sqrt{231}i}{6}-\frac{1}{2}\approx -0,5+2,533114026i$$

$$z=-\frac{\sqrt{231}i}{6}-\frac{1}{2}\approx -0,5-2,533114026i$$

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.

$$3z^{2}+3z+20=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 3 ni b va 20 ni c bilan almashtiring.

$$z=\frac{-3±\sqrt{3^{2}-4\times 3\times 20}}{2\times 3}$$

3 kvadratini chiqarish.

$$z=\frac{-3±\sqrt{9-4\times 3\times 20}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$z=\frac{-3±\sqrt{9-12\times 20}}{2\times 3}$$

-12 ni 20 marotabaga ko'paytirish.

$$z=\frac{-3±\sqrt{9-240}}{2\times 3}$$

9 ni -240 ga qo'shish.

$$z=\frac{-3±\sqrt{-231}}{2\times 3}$$

-231 ning kvadrat ildizini chiqarish.

$$z=\frac{-3±\sqrt{231}i}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$z=\frac{-3±\sqrt{231}i}{6}$$

z=\frac{-3±\sqrt{231}i}{6} tenglamasini yeching, bunda ± musbat. -3 ni i\sqrt{231} ga qo'shish.

$$z=\frac{-3+\sqrt{231}i}{6}$$

-3+i\sqrt{231} ni 6 ga bo'lish.

$$z=\frac{\sqrt{231}i}{6}-\frac{1}{2}$$

z=\frac{-3±\sqrt{231}i}{6} tenglamasini yeching, bunda ± manfiy. -3 dan i\sqrt{231} ni ayirish.

$$z=\frac{-\sqrt{231}i-3}{6}$$

-3-i\sqrt{231} ni 6 ga bo'lish.

$$z=-\frac{\sqrt{231}i}{6}-\frac{1}{2}$$

Tenglama yechildi.

$$z=\frac{\sqrt{231}i}{6}-\frac{1}{2} z=-\frac{\sqrt{231}i}{6}-\frac{1}{2}$$

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.

$$3z^{2}+3z+20=0$$

Tenglamaning ikkala tarafidan 20 ni ayirish.

$$3z^{2}+3z+20-20=-20$$

O‘zidan 20 ayirilsa 0 qoladi.

$$3z^{2}+3z=-20$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3z^{2}+3z}{3}=-\frac{20}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$z^{2}+\frac{3}{3}z=-\frac{20}{3}$$

3 ni 3 ga bo'lish.

$$z^{2}+z=-\frac{20}{3}$$

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.

$$z^{2}+z+\left(\frac{1}{2}\right)^{2}=-\frac{20}{3}+\left(\frac{1}{2}\right)^{2}$$

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

$$z^{2}+z+\frac{1}{4}=-\frac{20}{3}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{20}{3} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$z^{2}+z+\frac{1}{4}=-\frac{77}{12}$$

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.

$$\left(z+\frac{1}{2}\right)^{2}=-\frac{77}{12}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(z+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{77}{12}}$$

Qisqartirish.

$$z+\frac{1}{2}=\frac{\sqrt{231}i}{6} z+\frac{1}{2}=-\frac{\sqrt{231}i}{6}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$z=\frac{\sqrt{231}i}{6}-\frac{1}{2} z=-\frac{\sqrt{231}i}{6}-\frac{1}{2}$$