5x+3x^2=4 eching

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

x uchun yechish

$$x=\frac{\sqrt{73}-5}{6}\approx 0,590667291$$

$$x=\frac{-\sqrt{73}-5}{6}\approx -2,257333958$$

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.

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

Tenglamaning ikkala tarafidan 4 ni ayirish.

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

O‘zidan 4 ayirilsa 0 qoladi.

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

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.

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

5 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

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

-12 ni -4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25+48}}{2\times 3}$$

25 ni 48 ga qo'shish.

$$x=\frac{-5±\sqrt{73}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{73}}{6}$$

x=\frac{-5±\sqrt{73}}{6} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{73} ga qo'shish.

$$x=\frac{\sqrt{73}-5}{6}$$

x=\frac{-5±\sqrt{73}}{6} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{73} ni ayirish.

$$x=\frac{-\sqrt{73}-5}{6}$$

Tenglama yechildi.

$$x=\frac{\sqrt{73}-5}{6} x=\frac{-\sqrt{73}-5}{6}$$

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.

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

Ikki tarafini 3 ga bo‘ling.

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

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

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

\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.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{73}{36}$$

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.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{73}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{73}{36}}$$

Qisqartirish.

$$x+\frac{5}{6}=\frac{\sqrt{73}}{6} x+\frac{5}{6}=-\frac{\sqrt{73}}{6}$$

Tenglamaning ikkala tarafidan \frac{5}{6} ni ayirish.

$$x=\frac{\sqrt{73}-5}{6} x=\frac{-\sqrt{73}-5}{6}$$