2x^2+4x-4=0 eching

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

x uchun yechish (complex solution)

$$x=\sqrt{3}-1\approx 0,732050808$$

$$x=-\left(\sqrt{3}+1\right)\approx -2,732050808$$

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.

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

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

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

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-4\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-8\left(-4\right)}}{2\times 2}$$

-8 ni -4 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+32}}{2\times 2}$$

16 ni 32 ga qo'shish.

$$x=\frac{-4±\sqrt{48}}{2\times 2}$$

48 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4\sqrt{3}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±4\sqrt{3}}{4}$$

x=\frac{-4±4\sqrt{3}}{4} tenglamasini yeching, bunda ± musbat. -4 ni 4\sqrt{3} ga qo'shish.

$$x=\frac{4\sqrt{3}-4}{4}$$

-4+4\sqrt{3} ni 4 ga bo'lish.

$$x=\sqrt{3}-1$$

x=\frac{-4±4\sqrt{3}}{4} tenglamasini yeching, bunda ± manfiy. -4 dan 4\sqrt{3} ni ayirish.

$$x=\frac{-4\sqrt{3}-4}{4}$$

-4-4\sqrt{3} ni 4 ga bo'lish.

$$x=-\sqrt{3}-1$$

Tenglama yechildi.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

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.

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

4 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -4 ayirilsa 0 qoladi.

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

0 dan -4 ni ayirish.

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

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+4x}{2}=\frac{4}{2}$$

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

$$x^{2}+\frac{4}{2}x=\frac{4}{2}$$

4 ni 2 ga bo'lish.

$$x^{2}+2x=\frac{4}{2}$$

4 ni 2 ga bo'lish.

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

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+2x+1^{2}=2+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=2+1$$

2 ni 1 ga qo'shish.

$$x^{2}+2x+1=3$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{3}$$

Qisqartirish.

$$x+1=\sqrt{3} x+1=-\sqrt{3}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

x uchun yechish

$$x=\sqrt{3}-1\approx 0,732050808$$

$$x=-\sqrt{3}-1\approx -2,732050808$$

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.

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

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

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

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-4\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-8\left(-4\right)}}{2\times 2}$$

-8 ni -4 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+32}}{2\times 2}$$

16 ni 32 ga qo'shish.

$$x=\frac{-4±\sqrt{48}}{2\times 2}$$

48 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4\sqrt{3}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-4±4\sqrt{3}}{4}$$

x=\frac{-4±4\sqrt{3}}{4} tenglamasini yeching, bunda ± musbat. -4 ni 4\sqrt{3} ga qo'shish.

$$x=\frac{4\sqrt{3}-4}{4}$$

-4+4\sqrt{3} ni 4 ga bo'lish.

$$x=\sqrt{3}-1$$

x=\frac{-4±4\sqrt{3}}{4} tenglamasini yeching, bunda ± manfiy. -4 dan 4\sqrt{3} ni ayirish.

$$x=\frac{-4\sqrt{3}-4}{4}$$

-4-4\sqrt{3} ni 4 ga bo'lish.

$$x=-\sqrt{3}-1$$

Tenglama yechildi.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$

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.

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

4 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -4 ayirilsa 0 qoladi.

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

0 dan -4 ni ayirish.

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

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+4x}{2}=\frac{4}{2}$$

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

$$x^{2}+\frac{4}{2}x=\frac{4}{2}$$

4 ni 2 ga bo'lish.

$$x^{2}+2x=\frac{4}{2}$$

4 ni 2 ga bo'lish.

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

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+2x+1^{2}=2+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=2+1$$

2 ni 1 ga qo'shish.

$$x^{2}+2x+1=3$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{3}$$

Qisqartirish.

$$x+1=\sqrt{3} x+1=-\sqrt{3}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{3}-1 x=-\sqrt{3}-1$$