3x^2+5x+2=8 eching

$$3 { x }^{ 2 } +5x+2 = 8$$

x uchun yechish

$$x=\frac{\sqrt{97}-5}{6}\approx 0,808142967$$

$$x=\frac{-\sqrt{97}-5}{6}\approx -2,474809634$$

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.

$$3x^{2}+5x+2=8$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$3x^{2}+5x+2-8=8-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$3x^{2}+5x+2-8=0$$

2 dan 8 ni ayirish.

$$3x^{2}+5x-6=0$$

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 -6 ni c bilan almashtiring.

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

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 3\left(-6\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-12\left(-6\right)}}{2\times 3}$$

-12 ni -6 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25+72}}{2\times 3}$$

25 ni 72 ga qo'shish.

$$x=\frac{-5±\sqrt{97}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{97}}{6}$$

x=\frac{-5±\sqrt{97}}{6} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{97} ga qo'shish.

$$x=\frac{\sqrt{97}-5}{6}$$

x=\frac{-5±\sqrt{97}}{6} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{97} ni ayirish.

$$x=\frac{-\sqrt{97}-5}{6}$$

Tenglama yechildi.

$$x=\frac{\sqrt{97}-5}{6} x=\frac{-\sqrt{97}-5}{6}$$

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.

$$3x^{2}+5x+2=8$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$3x^{2}+5x+2-2=8-2$$

O‘zidan 2 ayirilsa 0 qoladi.

$$3x^{2}+5x=8-2$$

8 dan 2 ni ayirish.

$$3x^{2}+5x=6$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+5x}{3}=\frac{6}{3}$$

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

$$x^{2}+\frac{5}{3}x=\frac{6}{3}$$

6 ni 3 ga bo'lish.

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

\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.

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

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

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

2 ni \frac{25}{36} ga qo'shish.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{97}{36}$$

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.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{97}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{97}{36}}$$

Qisqartirish.

$$x+\frac{5}{6}=\frac{\sqrt{97}}{6} x+\frac{5}{6}=-\frac{\sqrt{97}}{6}$$

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

$$x=\frac{\sqrt{97}-5}{6} x=\frac{-\sqrt{97}-5}{6}$$