x^2+9=12x eching

$$x ^ { 2 } + 9 = 12 x$$

x uchun yechish

$$x=3\sqrt{3}+6\approx 11,196152423$$

$$x=6-3\sqrt{3}\approx 0,803847577$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 12x ni ayirish.

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

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.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -12 ni b va 9 ni c bilan almashtiring.

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

-12 kvadratini chiqarish.

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

-4 ni 9 marotabaga ko'paytirish.

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

144 ni -36 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{108}}{2}$$

108 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±6\sqrt{3}}{2}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±6\sqrt{3}}{2}$$

x=\frac{12±6\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. 12 ni 6\sqrt{3} ga qo'shish.

$$x=\frac{6\sqrt{3}+12}{2}$$

12+6\sqrt{3} ni 2 ga bo'lish.

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

x=\frac{12±6\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. 12 dan 6\sqrt{3} ni ayirish.

$$x=\frac{12-6\sqrt{3}}{2}$$

12-6\sqrt{3} ni 2 ga bo'lish.

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

Tenglama yechildi.

$$x=3\sqrt{3}+6 x=6-3\sqrt{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 12x ni ayirish.

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

Ikkala tarafdan 9 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

-12 ni bo‘lish, x shartining koeffitsienti, 2 ga -6 olish uchun. Keyin, -6 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-12x+\left(-6\right)^{2}=-9+\left(-6\right)^{2}$$

-6 kvadratini chiqarish.

$$x^{2}-12x+36=-9+36$$

-9 ni 36 ga qo'shish.

$$x^{2}-12x+36=27$$

x^{2}-12x+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.

$$\left(x-6\right)^{2}=27$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{27}$$

Qisqartirish.

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

6 ni tenglamaning ikkala tarafiga qo'shish.

$$x=3\sqrt{3}+6 x=6-3\sqrt{3}$$