12x^2-12x-6=0 eching

$$12 x ^ { 2 } - 12 x - 6 = 0$$

x uchun yechish

$$x = \frac{\sqrt{3} + 1}{2} \approx 1,366025404$$

$$x=\frac{1-\sqrt{3}}{2}\approx -0,366025404$$

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.

$$12x^{2}-12x-6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 12 ni a, -12 ni b va -6 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 12\left(-6\right)}}{2\times 12}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 12\left(-6\right)}}{2\times 12}$$

-4 ni 12 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-48\left(-6\right)}}{2\times 12}$$

-48 ni -6 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+288}}{2\times 12}$$

144 ni 288 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{432}}{2\times 12}$$

432 ning kvadrat ildizini chiqarish.

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

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±12\sqrt{3}}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$x=\frac{12±12\sqrt{3}}{24}$$

x=\frac{12±12\sqrt{3}}{24} tenglamasini yeching, bunda ± musbat. 12 ni 12\sqrt{3} ga qo'shish.

$$x=\frac{12\sqrt{3}+12}{24}$$

12+12\sqrt{3} ni 24 ga bo'lish.

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

x=\frac{12±12\sqrt{3}}{24} tenglamasini yeching, bunda ± manfiy. 12 dan 12\sqrt{3} ni ayirish.

$$x=\frac{12-12\sqrt{3}}{24}$$

12-12\sqrt{3} ni 24 ga bo'lish.

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

Tenglama yechildi.

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

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.

$$12x^{2}-12x-6=0$$

6 ni tenglamaning ikkala tarafiga qo'shish.

$$12x^{2}-12x-6-\left(-6\right)=-\left(-6\right)$$

O‘zidan -6 ayirilsa 0 qoladi.

$$12x^{2}-12x=-\left(-6\right)$$

0 dan -6 ni ayirish.

$$12x^{2}-12x=6$$

Ikki tarafini 12 ga bo‘ling.

$$\frac{12x^{2}-12x}{12}=\frac{6}{12}$$

12 ga bo'lish 12 ga ko'paytirishni bekor qiladi.

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

-12 ni 12 ga bo'lish.

$$x^{2}-x=\frac{6}{12}$$

\frac{6}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-x=\frac{1}{2}$$

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

$$x^{2}-x+\frac{1}{4}=\frac{1}{2}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-x+\frac{1}{4}=\frac{3}{4}$$

x^{2}-x+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{3}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Qisqartirish.

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

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

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