2x^2-12x-1=0 eching

$$2 { x }^{ 2 } -12x-1=0$$

x uchun yechish

$$x=\frac{\sqrt{38}}{2}+3\approx 6,082207001$$

$$x=-\frac{\sqrt{38}}{2}+3\approx -0,082207001$$

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.

$$2x^{2}-12x-1=0$$

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

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\left(-1\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\left(-1\right)}}{2\times 2}$$

-8 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+8}}{2\times 2}$$

144 ni 8 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{152}}{2\times 2}$$

152 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±2\sqrt{38}}{2\times 2}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±2\sqrt{38}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{12±2\sqrt{38}}{4}$$

x=\frac{12±2\sqrt{38}}{4} tenglamasini yeching, bunda ± musbat. 12 ni 2\sqrt{38} ga qo'shish.

$$x=\frac{2\sqrt{38}+12}{4}$$

12+2\sqrt{38} ni 4 ga bo'lish.

$$x=\frac{\sqrt{38}}{2}+3$$

x=\frac{12±2\sqrt{38}}{4} tenglamasini yeching, bunda ± manfiy. 12 dan 2\sqrt{38} ni ayirish.

$$x=\frac{12-2\sqrt{38}}{4}$$

12-2\sqrt{38} ni 4 ga bo'lish.

$$x=-\frac{\sqrt{38}}{2}+3$$

Tenglama yechildi.

$$x=\frac{\sqrt{38}}{2}+3 x=-\frac{\sqrt{38}}{2}+3$$

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.

$$2x^{2}-12x-1=0$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}-12x-1-\left(-1\right)=-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$2x^{2}-12x=-\left(-1\right)$$

0 dan -1 ni ayirish.

$$2x^{2}-12x=1$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-12x}{2}=\frac{1}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{12}{2}\right)x=\frac{1}{2}$$

-12 ni 2 ga bo'lish.

$$x^{2}-6x=\frac{1}{2}$$

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

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

-3 kvadratini chiqarish.

$$x^{2}-6x+9=\frac{1}{2}+9$$

\frac{1}{2} ni 9 ga qo'shish.

$$x^{2}-6x+9=\frac{19}{2}$$

x^{2}-6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-3\right)^{2}=\frac{19}{2}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{\frac{19}{2}}$$

Qisqartirish.

$$x-3=\frac{\sqrt{38}}{2} x-3=-\frac{\sqrt{38}}{2}$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{38}}{2}+3 x=-\frac{\sqrt{38}}{2}+3$$