3x^2+2x+5=18 eching

$$3 x ^ { 2 } + 2 x + 5 = 18$$

x uchun yechish

$$x = \frac{2 \sqrt{10} - 1}{3} \approx 1,774851773$$

$$x=\frac{-2\sqrt{10}-1}{3}\approx -2,44151844$$

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.

$$3x^{2}+2x+5=18$$

Tenglamaning ikkala tarafidan 18 ni ayirish.

$$3x^{2}+2x+5-18=18-18$$

O‘zidan 18 ayirilsa 0 qoladi.

$$3x^{2}+2x+5-18=0$$

5 dan 18 ni ayirish.

$$3x^{2}+2x-13=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 2 ni b va -13 ni c bilan almashtiring.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\left(-13\right)}}{2\times 3}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\times 3\left(-13\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-12\left(-13\right)}}{2\times 3}$$

-12 ni -13 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+156}}{2\times 3}$$

4 ni 156 ga qo'shish.

$$x=\frac{-2±\sqrt{160}}{2\times 3}$$

160 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±4\sqrt{10}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-2±4\sqrt{10}}{6}$$

x=\frac{-2±4\sqrt{10}}{6} tenglamasini yeching, bunda ± musbat. -2 ni 4\sqrt{10} ga qo'shish.

$$x=\frac{4\sqrt{10}-2}{6}$$

-2+4\sqrt{10} ni 6 ga bo'lish.

$$x=\frac{2\sqrt{10}-1}{3}$$

x=\frac{-2±4\sqrt{10}}{6} tenglamasini yeching, bunda ± manfiy. -2 dan 4\sqrt{10} ni ayirish.

$$x=\frac{-4\sqrt{10}-2}{6}$$

-2-4\sqrt{10} ni 6 ga bo'lish.

$$x=\frac{-2\sqrt{10}-1}{3}$$

Tenglama yechildi.

$$x=\frac{2\sqrt{10}-1}{3} x=\frac{-2\sqrt{10}-1}{3}$$

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.

$$3x^{2}+2x+5=18$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$3x^{2}+2x+5-5=18-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$3x^{2}+2x=18-5$$

18 dan 5 ni ayirish.

$$3x^{2}+2x=13$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+2x}{3}=\frac{13}{3}$$

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

$$x^{2}+\frac{2}{3}x=\frac{13}{3}$$

\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{3} olish uchun. Keyin, \frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{13}{3}+\left(\frac{1}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{3} kvadratini chiqarish.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{13}{3}+\frac{1}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{13}{3} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{40}{9}$$

x^{2}+\frac{2}{3}x+\frac{1}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{40}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{40}{9}}$$

Qisqartirish.

$$x+\frac{1}{3}=\frac{2\sqrt{10}}{3} x+\frac{1}{3}=-\frac{2\sqrt{10}}{3}$$

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

$$x=\frac{2\sqrt{10}-1}{3} x=\frac{-2\sqrt{10}-1}{3}$$