-3x^2+5x-4=0 eching

$$- 3 x ^ { 2 } + 5 x - 4 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-\sqrt{23}i+5}{6}\approx 0,833333333-0,799305254i$$

$$x=\frac{5+\sqrt{23}i}{6}\approx 0,833333333+0,799305254i$$

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.

$$-3x^{2}+5x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -3 ni a, 5 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25+12\left(-4\right)}}{2\left(-3\right)}$$

12 ni -4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-48}}{2\left(-3\right)}$$

25 ni -48 ga qo'shish.

$$x=\frac{-5±\sqrt{-23}}{2\left(-3\right)}$$

-23 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±\sqrt{23}i}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{23}i}{-6}$$

x=\frac{-5±\sqrt{23}i}{-6} tenglamasini yeching, bunda ± musbat. -5 ni i\sqrt{23} ga qo'shish.

$$x=\frac{-5+\sqrt{23}i}{-6}$$

-5+i\sqrt{23} ni -6 ga bo'lish.

$$x=\frac{-\sqrt{23}i+5}{6}$$

x=\frac{-5±\sqrt{23}i}{-6} tenglamasini yeching, bunda ± manfiy. -5 dan i\sqrt{23} ni ayirish.

$$x=\frac{-\sqrt{23}i-5}{-6}$$

-5-i\sqrt{23} ni -6 ga bo'lish.

$$x=\frac{5+\sqrt{23}i}{6}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{23}i+5}{6} x=\frac{5+\sqrt{23}i}{6}$$

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.

$$-3x^{2}+5x-4=0$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$-3x^{2}+5x-4-\left(-4\right)=-\left(-4\right)$$

O‘zidan -4 ayirilsa 0 qoladi.

$$-3x^{2}+5x=-\left(-4\right)$$

0 dan -4 ni ayirish.

$$-3x^{2}+5x=4$$

Ikki tarafini -3 ga bo‘ling.

$$\frac{-3x^{2}+5x}{-3}=\frac{4}{-3}$$

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

$$x^{2}+\frac{5}{-3}x=\frac{4}{-3}$$

5 ni -3 ga bo'lish.

$$x^{2}-\frac{5}{3}x=\frac{4}{-3}$$

4 ni -3 ga bo'lish.

$$x^{2}-\frac{5}{3}x=-\frac{4}{3}$$

-\frac{5}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{6} olish uchun. Keyin, -\frac{5}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{4}{3}+\left(-\frac{5}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{6} kvadratini chiqarish.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{4}{3}+\frac{25}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{4}{3} ni \frac{25}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{23}{36}$$

x^{2}-\frac{5}{3}x+\frac{25}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{6}\right)^{2}=-\frac{23}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{-\frac{23}{36}}$$

Qisqartirish.

$$x-\frac{5}{6}=\frac{\sqrt{23}i}{6} x-\frac{5}{6}=-\frac{\sqrt{23}i}{6}$$

\frac{5}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{5+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i+5}{6}$$