x^2-12x-5=-2 eching

$${ x }^{ 2 } -12x-5=-2$$

x uchun yechish

$$x=\sqrt{39}+6\approx 12,244997998$$

$$x=6-\sqrt{39}\approx -0,244997998$$

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.

$$x^{2}-12x-5=-2$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}-12x-5-\left(-2\right)=-2-\left(-2\right)$$

O‘zidan -2 ayirilsa 0 qoladi.

$$x^{2}-12x-5-\left(-2\right)=0$$

-5 dan -2 ni ayirish.

$$x^{2}-12x-3=0$$

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 -3 ni c bilan almashtiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-3\right)}}{2}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-3\right)}}{2}$$

-4 ni -3 marotabaga ko'paytirish.

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

144 ni 12 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{156}}{2}$$

156 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±2\sqrt{39}}{2}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±2\sqrt{39}}{2}$$

x=\frac{12±2\sqrt{39}}{2} tenglamasini yeching, bunda ± musbat. 12 ni 2\sqrt{39} ga qo'shish.

$$x=\frac{2\sqrt{39}+12}{2}$$

12+2\sqrt{39} ni 2 ga bo'lish.

$$x=\sqrt{39}+6$$

x=\frac{12±2\sqrt{39}}{2} tenglamasini yeching, bunda ± manfiy. 12 dan 2\sqrt{39} ni ayirish.

$$x=\frac{12-2\sqrt{39}}{2}$$

12-2\sqrt{39} ni 2 ga bo'lish.

$$x=6-\sqrt{39}$$

Tenglama yechildi.

$$x=\sqrt{39}+6 x=6-\sqrt{39}$$

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.

$$x^{2}-12x-5=-2$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}-12x-5-\left(-5\right)=-2-\left(-5\right)$$

O‘zidan -5 ayirilsa 0 qoladi.

$$x^{2}-12x=-2-\left(-5\right)$$

-2 dan -5 ni ayirish.

$$x^{2}-12x=3$$

-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.

$$x^{2}-12x+\left(-6\right)^{2}=3+\left(-6\right)^{2}$$

-6 kvadratini chiqarish.

$$x^{2}-12x+36=3+36$$

3 ni 36 ga qo'shish.

$$x^{2}-12x+36=39$$

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.

$$\left(x-6\right)^{2}=39$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{39}$$

Qisqartirish.

$$x-6=\sqrt{39} x-6=-\sqrt{39}$$

6 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{39}+6 x=6-\sqrt{39}$$