9^2+2x^2=5x eching

$$9 ^ { 2 } + 2 x ^ { 2 } = 5 x$$

x uchun yechish (complex solution)

$$x=\frac{5+\sqrt{623}i}{4}\approx 1,25+6,239991987i$$

$$x=\frac{-\sqrt{623}i+5}{4}\approx 1,25-6,239991987i$$

Kvadrat tenglamalaridan foydalanish qoidalari

2 daraja ko‘rsatkichini 9 ga hisoblang va 81 ni qiymatni oling.

$$81+2x^{2}=5x$$

Ikkala tarafdan 5x ni ayirish.

$$81+2x^{2}-5x=0$$

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.

$$2x^{2}-5x+81=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -5 ni b va 81 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\times 81}}{2\times 2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\times 81}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\times 81}}{2\times 2}$$

-8 ni 81 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-648}}{2\times 2}$$

25 ni -648 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{-623}}{2\times 2}$$

-623 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{623}i}{2\times 2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{623}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{5±\sqrt{623}i}{4}$$

x=\frac{5±\sqrt{623}i}{4} tenglamasini yeching, bunda ± musbat. 5 ni i\sqrt{623} ga qo'shish.

$$x=\frac{5+\sqrt{623}i}{4}$$

x=\frac{5±\sqrt{623}i}{4} tenglamasini yeching, bunda ± manfiy. 5 dan i\sqrt{623} ni ayirish.

$$x=\frac{-\sqrt{623}i+5}{4}$$

Tenglama yechildi.

$$x=\frac{5+\sqrt{623}i}{4} x=\frac{-\sqrt{623}i+5}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

2 daraja ko‘rsatkichini 9 ga hisoblang va 81 ni qiymatni oling.

$$81+2x^{2}=5x$$

Ikkala tarafdan 5x ni ayirish.

$$81+2x^{2}-5x=0$$

Ikkala tarafdan 81 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$2x^{2}-5x=-81$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-5x}{2}=-\frac{81}{2}$$

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

$$x^{2}-\frac{5}{2}x=-\frac{81}{2}$$

-\frac{5}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{4} olish uchun. Keyin, -\frac{5}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{81}{2}+\left(-\frac{5}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{4} kvadratini chiqarish.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{81}{2}+\frac{25}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{81}{2} ni \frac{25}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{623}{16}$$

x^{2}-\frac{5}{2}x+\frac{25}{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.

$$\left(x-\frac{5}{4}\right)^{2}=-\frac{623}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{-\frac{623}{16}}$$

Qisqartirish.

$$x-\frac{5}{4}=\frac{\sqrt{623}i}{4} x-\frac{5}{4}=-\frac{\sqrt{623}i}{4}$$

\frac{5}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{5+\sqrt{623}i}{4} x=\frac{-\sqrt{623}i+5}{4}$$