x^2-12x=4 eching

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

x uchun yechish

$$x=2\sqrt{10}+6\approx 12,32455532$$

$$x=6-2\sqrt{10}\approx -0,32455532$$

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.

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

Tenglamaning ikkala tarafidan 4 ni ayirish.

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

O‘zidan 4 ayirilsa 0 qoladi.

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

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 -4 ni c bilan almashtiring.

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

-12 kvadratini chiqarish.

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

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+16}}{2}$$

144 ni 16 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{160}}{2}$$

160 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±4\sqrt{10}}{2}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±4\sqrt{10}}{2}$$

x=\frac{12±4\sqrt{10}}{2} tenglamasini yeching, bunda ± musbat. 12 ni 4\sqrt{10} ga qo'shish.

$$x=\frac{4\sqrt{10}+12}{2}$$

12+4\sqrt{10} ni 2 ga bo'lish.

$$x=2\sqrt{10}+6$$

x=\frac{12±4\sqrt{10}}{2} tenglamasini yeching, bunda ± manfiy. 12 dan 4\sqrt{10} ni ayirish.

$$x=\frac{12-4\sqrt{10}}{2}$$

12-4\sqrt{10} ni 2 ga bo'lish.

$$x=6-2\sqrt{10}$$

Tenglama yechildi.

$$x=2\sqrt{10}+6 x=6-2\sqrt{10}$$

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.

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

-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.

$$x^{2}-12x+\left(-6\right)^{2}=4+\left(-6\right)^{2}$$

-6 kvadratini chiqarish.

$$x^{2}-12x+36=4+36$$

4 ni 36 ga qo'shish.

$$x^{2}-12x+36=40$$

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.

$$\left(x-6\right)^{2}=40$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{40}$$

Qisqartirish.

$$x-6=2\sqrt{10} x-6=-2\sqrt{10}$$

6 ni tenglamaning ikkala tarafiga qo'shish.

$$x=2\sqrt{10}+6 x=6-2\sqrt{10}$$