9x^2+9x=1 eching

$$9 x ^ { 2 } + 9 x = 1$$

x uchun yechish

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2}\approx 0,100925213$$

$$x=-\frac{\sqrt{13}}{6}-\frac{1}{2}\approx -1,100925213$$

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.

$$9x^{2}+9x=1$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$9x^{2}+9x-1=1-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$9x^{2}+9x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, 9 ni b va -1 ni c bilan almashtiring.

$$x=\frac{-9±\sqrt{9^{2}-4\times 9\left(-1\right)}}{2\times 9}$$

9 kvadratini chiqarish.

$$x=\frac{-9±\sqrt{81-4\times 9\left(-1\right)}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81-36\left(-1\right)}}{2\times 9}$$

-36 ni -1 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81+36}}{2\times 9}$$

81 ni 36 ga qo'shish.

$$x=\frac{-9±\sqrt{117}}{2\times 9}$$

117 ning kvadrat ildizini chiqarish.

$$x=\frac{-9±3\sqrt{13}}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-9±3\sqrt{13}}{18}$$

x=\frac{-9±3\sqrt{13}}{18} tenglamasini yeching, bunda ± musbat. -9 ni 3\sqrt{13} ga qo'shish.

$$x=\frac{3\sqrt{13}-9}{18}$$

-9+3\sqrt{13} ni 18 ga bo'lish.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2}$$

x=\frac{-9±3\sqrt{13}}{18} tenglamasini yeching, bunda ± manfiy. -9 dan 3\sqrt{13} ni ayirish.

$$x=\frac{-3\sqrt{13}-9}{18}$$

-9-3\sqrt{13} ni 18 ga bo'lish.

$$x=-\frac{\sqrt{13}}{6}-\frac{1}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2} x=-\frac{\sqrt{13}}{6}-\frac{1}{2}$$

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.

$$9x^{2}+9x=1$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}+9x}{9}=\frac{1}{9}$$

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

$$x^{2}+\frac{9}{9}x=\frac{1}{9}$$

9 ni 9 ga bo'lish.

$$x^{2}+x=\frac{1}{9}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{1}{9}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{1}{9}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{9} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+x+\frac{1}{4}=\frac{13}{36}$$

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.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{13}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Qisqartirish.

$$x+\frac{1}{2}=\frac{\sqrt{13}}{6} x+\frac{1}{2}=-\frac{\sqrt{13}}{6}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{2} x=-\frac{\sqrt{13}}{6}-\frac{1}{2}$$