7x^2+12x+4=0 eching

$$7 { x }^{ 2 } +12x+4 = 0$$

x uchun yechish

$$x=\frac{2\sqrt{2}-6}{7}\approx -0,453081839$$

$$x=\frac{-2\sqrt{2}-6}{7}\approx -1,261203875$$

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.

$$7x^{2}+12x+4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 7 ni a, 12 ni b va 4 ni c bilan almashtiring.

$$x=\frac{-12±\sqrt{12^{2}-4\times 7\times 4}}{2\times 7}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 7\times 4}}{2\times 7}$$

-4 ni 7 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-28\times 4}}{2\times 7}$$

-28 ni 4 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-112}}{2\times 7}$$

144 ni -112 ga qo'shish.

$$x=\frac{-12±\sqrt{32}}{2\times 7}$$

32 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±4\sqrt{2}}{2\times 7}$$

2 ni 7 marotabaga ko'paytirish.

$$x=\frac{-12±4\sqrt{2}}{14}$$

x=\frac{-12±4\sqrt{2}}{14} tenglamasini yeching, bunda ± musbat. -12 ni 4\sqrt{2} ga qo'shish.

$$x=\frac{4\sqrt{2}-12}{14}$$

-12+4\sqrt{2} ni 14 ga bo'lish.

$$x=\frac{2\sqrt{2}-6}{7}$$

x=\frac{-12±4\sqrt{2}}{14} tenglamasini yeching, bunda ± manfiy. -12 dan 4\sqrt{2} ni ayirish.

$$x=\frac{-4\sqrt{2}-12}{14}$$

-12-4\sqrt{2} ni 14 ga bo'lish.

$$x=\frac{-2\sqrt{2}-6}{7}$$

Tenglama yechildi.

$$x=\frac{2\sqrt{2}-6}{7} x=\frac{-2\sqrt{2}-6}{7}$$

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.

$$7x^{2}+12x+4=0$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$7x^{2}+12x+4-4=-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$7x^{2}+12x=-4$$

Ikki tarafini 7 ga bo‘ling.

$$\frac{7x^{2}+12x}{7}=-\frac{4}{7}$$

7 ga bo'lish 7 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{12}{7}x=-\frac{4}{7}$$

\frac{12}{7} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{6}{7} olish uchun. Keyin, \frac{6}{7} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{12}{7}x+\left(\frac{6}{7}\right)^{2}=-\frac{4}{7}+\left(\frac{6}{7}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{6}{7} kvadratini chiqarish.

$$x^{2}+\frac{12}{7}x+\frac{36}{49}=-\frac{4}{7}+\frac{36}{49}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{4}{7} ni \frac{36}{49} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{12}{7}x+\frac{36}{49}=\frac{8}{49}$$

x^{2}+\frac{12}{7}x+\frac{36}{49} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{6}{7}\right)^{2}=\frac{8}{49}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{6}{7}\right)^{2}}=\sqrt{\frac{8}{49}}$$

Qisqartirish.

$$x+\frac{6}{7}=\frac{2\sqrt{2}}{7} x+\frac{6}{7}=-\frac{2\sqrt{2}}{7}$$

Tenglamaning ikkala tarafidan \frac{6}{7} ni ayirish.

$$x=\frac{2\sqrt{2}-6}{7} x=\frac{-2\sqrt{2}-6}{7}$$