x^2+(2-x)^2=(2x)^2 eching

$$x ^ { 2 } + ( 2 - x ) ^ { 2 } = ( 2 x ) ^ { 2 }$$

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2-x\right)^{2} kengaytirilishi uchun ishlating.

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

2x^{2} ni olish uchun x^{2} va x^{2} ni birlashtirish.

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

\left(2x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

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

Ikkala tarafdan 4x^{2} ni ayirish.

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

-2x^{2} ni olish uchun 2x^{2} va -4x^{2} ni birlashtirish.

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

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.

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

-4 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

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

8 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\left(-2\right)}$$

16 ni 32 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2\left(-2\right)}$$

48 ning kvadrat ildizini chiqarish.

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

-4 ning teskarisi 4 ga teng.

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

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.

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

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

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

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

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2-x\right)^{2} kengaytirilishi uchun ishlating.

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

2x^{2} ni olish uchun x^{2} va x^{2} ni birlashtirish.

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

\left(2x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

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

Ikkala tarafdan 4x^{2} ni ayirish.

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

-2x^{2} ni olish uchun 2x^{2} va -4x^{2} ni birlashtirish.

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

Ikkala tarafdan 4 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

Ikki tarafini -2 ga bo‘ling.

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

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

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

-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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2-x\right)^{2} kengaytirilishi uchun ishlating.

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

2x^{2} ni olish uchun x^{2} va x^{2} ni birlashtirish.

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

\left(2x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

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

Ikkala tarafdan 4x^{2} ni ayirish.

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

-2x^{2} ni olish uchun 2x^{2} va -4x^{2} ni birlashtirish.

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

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.

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

-4 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

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

8 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+32}}{2\left(-2\right)}$$

16 ni 32 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{48}}{2\left(-2\right)}$$

48 ning kvadrat ildizini chiqarish.

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

-4 ning teskarisi 4 ga teng.

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

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.

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

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

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

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

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2-x\right)^{2} kengaytirilishi uchun ishlating.

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

2x^{2} ni olish uchun x^{2} va x^{2} ni birlashtirish.

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

\left(2x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

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

Ikkala tarafdan 4x^{2} ni ayirish.

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

-2x^{2} ni olish uchun 2x^{2} va -4x^{2} ni birlashtirish.

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

Ikkala tarafdan 4 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

Ikki tarafini -2 ga bo‘ling.

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

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

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

-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.

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