3x^2+x=11 eching

$$3 x ^ { 2 } + x = 11$$

x uchun yechish

$$x = \frac{\sqrt{133} - 1}{6} \approx 1,755427099$$

$$x=\frac{-\sqrt{133}-1}{6}\approx -2,088760432$$

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.

$$3x^{2}+x=11$$

Tenglamaning ikkala tarafidan 11 ni ayirish.

$$3x^{2}+x-11=11-11$$

O‘zidan 11 ayirilsa 0 qoladi.

$$3x^{2}+x-11=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 1 ni b va -11 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-11\right)}}{2\times 3}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-11\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-12\left(-11\right)}}{2\times 3}$$

-12 ni -11 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+132}}{2\times 3}$$

1 ni 132 ga qo'shish.

$$x=\frac{-1±\sqrt{133}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{133}}{6}$$

x=\frac{-1±\sqrt{133}}{6} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{133} ga qo'shish.

$$x=\frac{\sqrt{133}-1}{6}$$

x=\frac{-1±\sqrt{133}}{6} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{133} ni ayirish.

$$x=\frac{-\sqrt{133}-1}{6}$$

Tenglama yechildi.

$$x=\frac{\sqrt{133}-1}{6} x=\frac{-\sqrt{133}-1}{6}$$

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.

$$3x^{2}+x=11$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+x}{3}=\frac{11}{3}$$

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

$$x^{2}+\frac{1}{3}x=\frac{11}{3}$$

\frac{1}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{6} olish uchun. Keyin, \frac{1}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{11}{3}+\left(\frac{1}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{6} kvadratini chiqarish.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{11}{3}+\frac{1}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{11}{3} ni \frac{1}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{133}{36}$$

x^{2}+\frac{1}{3}x+\frac{1}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{133}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{133}{36}}$$

Qisqartirish.

$$x+\frac{1}{6}=\frac{\sqrt{133}}{6} x+\frac{1}{6}=-\frac{\sqrt{133}}{6}$$

Tenglamaning ikkala tarafidan \frac{1}{6} ni ayirish.

$$x=\frac{\sqrt{133}-1}{6} x=\frac{-\sqrt{133}-1}{6}$$