9x^2-12x-4=0 eching

$$9 x ^ { 2 } - 12 x - 4 = 0$$

x uchun yechish

$$x = \frac{2 \sqrt{2} + 2}{3} \approx 1,609475708$$

$$x=\frac{2-2\sqrt{2}}{3}\approx -0,276142375$$

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.

$$9x^{2}-12x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, -12 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\left(-4\right)}}{2\times 9}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\left(-4\right)}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-36\left(-4\right)}}{2\times 9}$$

-36 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+144}}{2\times 9}$$

144 ni 144 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{288}}{2\times 9}$$

288 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±12\sqrt{2}}{2\times 9}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±12\sqrt{2}}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{12±12\sqrt{2}}{18}$$

x=\frac{12±12\sqrt{2}}{18} tenglamasini yeching, bunda ± musbat. 12 ni 12\sqrt{2} ga qo'shish.

$$x=\frac{12\sqrt{2}+12}{18}$$

12+12\sqrt{2} ni 18 ga bo'lish.

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

x=\frac{12±12\sqrt{2}}{18} tenglamasini yeching, bunda ± manfiy. 12 dan 12\sqrt{2} ni ayirish.

$$x=\frac{12-12\sqrt{2}}{18}$$

12-12\sqrt{2} ni 18 ga bo'lish.

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

Tenglama yechildi.

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

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.

$$9x^{2}-12x-4=0$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$9x^{2}-12x-4-\left(-4\right)=-\left(-4\right)$$

O‘zidan -4 ayirilsa 0 qoladi.

$$9x^{2}-12x=-\left(-4\right)$$

0 dan -4 ni ayirish.

$$9x^{2}-12x=4$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}-12x}{9}=\frac{4}{9}$$

9 ga bo'lish 9 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{12}{9}\right)x=\frac{4}{9}$$

\frac{-12}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{4}{3}x=\frac{4}{9}$$

-\frac{4}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{2}{3} olish uchun. Keyin, -\frac{2}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{4}{9}+\left(-\frac{2}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{2}{3} kvadratini chiqarish.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4+4}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{4}{9} ni \frac{4}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{8}{9}$$

x^{2}-\frac{4}{3}x+\frac{4}{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.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{8}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{8}{9}}$$

Qisqartirish.

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

\frac{2}{3} ni tenglamaning ikkala tarafiga qo'shish.

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