3x^2-3x+4=26 eching

$$3 x ^ { 2 } - 3 x + 4 = 26$$

x uchun yechish

$$x=\frac{\sqrt{273}}{6}+\frac{1}{2}\approx 3,253785274$$

$$x=-\frac{\sqrt{273}}{6}+\frac{1}{2}\approx -2,253785274$$

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.

$$3x^{2}-3x+4=26$$

Tenglamaning ikkala tarafidan 26 ni ayirish.

$$3x^{2}-3x+4-26=26-26$$

O‘zidan 26 ayirilsa 0 qoladi.

$$3x^{2}-3x+4-26=0$$

4 dan 26 ni ayirish.

$$3x^{2}-3x-22=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -3 ni b va -22 ni c bilan almashtiring.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-22\right)}}{2\times 3}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-22\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-12\left(-22\right)}}{2\times 3}$$

-12 ni -22 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+264}}{2\times 3}$$

9 ni 264 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{273}}{2\times 3}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±\sqrt{273}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{3±\sqrt{273}}{6}$$

x=\frac{3±\sqrt{273}}{6} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{273} ga qo'shish.

$$x=\frac{\sqrt{273}+3}{6}$$

3+\sqrt{273} ni 6 ga bo'lish.

$$x=\frac{\sqrt{273}}{6}+\frac{1}{2}$$

x=\frac{3±\sqrt{273}}{6} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{273} ni ayirish.

$$x=\frac{3-\sqrt{273}}{6}$$

3-\sqrt{273} ni 6 ga bo'lish.

$$x=-\frac{\sqrt{273}}{6}+\frac{1}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{273}}{6}+\frac{1}{2} x=-\frac{\sqrt{273}}{6}+\frac{1}{2}$$

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.

$$3x^{2}-3x+4=26$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$3x^{2}-3x+4-4=26-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$3x^{2}-3x=26-4$$

26 dan 4 ni ayirish.

$$3x^{2}-3x=22$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-3x}{3}=\frac{22}{3}$$

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

$$x^{2}+\left(-\frac{3}{3}\right)x=\frac{22}{3}$$

-3 ni 3 ga bo'lish.

$$x^{2}-x=\frac{22}{3}$$

-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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{22}{3}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{22}{3}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{22}{3} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-x+\frac{1}{4}=\frac{91}{12}$$

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.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{91}{12}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{91}{12}}$$

Qisqartirish.

$$x-\frac{1}{2}=\frac{\sqrt{273}}{6} x-\frac{1}{2}=-\frac{\sqrt{273}}{6}$$

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

$$x=\frac{\sqrt{273}}{6}+\frac{1}{2} x=-\frac{\sqrt{273}}{6}+\frac{1}{2}$$