x^2+6x-2=2 eching

$$x ^ { 2 } + 6 x - 2 = 2$$

x uchun yechish (complex solution)

$$x=\sqrt{13}-3\approx 0,605551275$$

$$x=-\left(\sqrt{13}+3\right)\approx -6,605551275$$

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.

$$x^{2}+6x-2=2$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$x^{2}+6x-2-2=2-2$$

O‘zidan 2 ayirilsa 0 qoladi.

$$x^{2}+6x-2-2=0$$

-2 dan 2 ni ayirish.

$$x^{2}+6x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 6 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-4\right)}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-4\right)}}{2}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+16}}{2}$$

36 ni 16 ga qo'shish.

$$x=\frac{-6±\sqrt{52}}{2}$$

52 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{13}}{2}$$

x=\frac{-6±2\sqrt{13}}{2} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{13} ga qo'shish.

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

-6+2\sqrt{13} ni 2 ga bo'lish.

$$x=\sqrt{13}-3$$

x=\frac{-6±2\sqrt{13}}{2} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{13} ni ayirish.

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

-6-2\sqrt{13} ni 2 ga bo'lish.

$$x=-\sqrt{13}-3$$

Tenglama yechildi.

$$x=\sqrt{13}-3 x=-\sqrt{13}-3$$

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.

$$x^{2}+6x-2=2$$

2 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -2 ayirilsa 0 qoladi.

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

2 dan -2 ni ayirish.

$$x^{2}+6x=4$$

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.

$$x^{2}+6x+3^{2}=4+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=4+9$$

4 ni 9 ga qo'shish.

$$x^{2}+6x+9=13$$

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.

$$\left(x+3\right)^{2}=13$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{13}$$

Qisqartirish.

$$x+3=\sqrt{13} x+3=-\sqrt{13}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=\sqrt{13}-3 x=-\sqrt{13}-3$$

x uchun yechish

$$x=\sqrt{13}-3\approx 0,605551275$$

$$x=-\sqrt{13}-3\approx -6,605551275$$

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.

$$x^{2}+6x-2=2$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$x^{2}+6x-2-2=2-2$$

O‘zidan 2 ayirilsa 0 qoladi.

$$x^{2}+6x-2-2=0$$

-2 dan 2 ni ayirish.

$$x^{2}+6x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 6 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-4\right)}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-4\right)}}{2}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+16}}{2}$$

36 ni 16 ga qo'shish.

$$x=\frac{-6±\sqrt{52}}{2}$$

52 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{13}}{2}$$

x=\frac{-6±2\sqrt{13}}{2} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{13} ga qo'shish.

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

-6+2\sqrt{13} ni 2 ga bo'lish.

$$x=\sqrt{13}-3$$

x=\frac{-6±2\sqrt{13}}{2} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{13} ni ayirish.

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

-6-2\sqrt{13} ni 2 ga bo'lish.

$$x=-\sqrt{13}-3$$

Tenglama yechildi.

$$x=\sqrt{13}-3 x=-\sqrt{13}-3$$

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.

$$x^{2}+6x-2=2$$

2 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -2 ayirilsa 0 qoladi.

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

2 dan -2 ni ayirish.

$$x^{2}+6x=4$$

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.

$$x^{2}+6x+3^{2}=4+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=4+9$$

4 ni 9 ga qo'shish.

$$x^{2}+6x+9=13$$

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.

$$\left(x+3\right)^{2}=13$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{13}$$

Qisqartirish.

$$x+3=\sqrt{13} x+3=-\sqrt{13}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=\sqrt{13}-3 x=-\sqrt{13}-3$$