x^2-(2x-1)^2/2=1 eching

$$x ^ { 2 } - \frac { ( 2 x - 1 ) ^ { 2 } } { 2 } = 1$$

x uchun yechish (complex solution)

$$x=\frac{\sqrt{2}i}{2}+1\approx 1+0,707106781i$$

$$x=-\frac{\sqrt{2}i}{2}+1\approx 1-0,707106781i$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tenglamaning ikkala tarafini 2 ga ko'paytirish.

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2x-1\right)^{2} kengaytirilishi uchun ishlating.

$$2x^{2}-\left(4x^{2}-4x+1\right)=2$$

4x^{2}-4x+1 teskarisini topish uchun har birining teskarisini toping.

$$2x^{2}-4x^{2}+4x-1=2$$

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

$$-2x^{2}+4x-1=2$$

Ikkala tarafdan 2 ni ayirish.

$$-2x^{2}+4x-1-2=0$$

-3 olish uchun -1 dan 2 ni ayirish.

$$-2x^{2}+4x-3=0$$

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 -3 ni c bilan almashtiring.

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

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+8\left(-3\right)}}{2\left(-2\right)}$$

8 ni -3 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-24}}{2\left(-2\right)}$$

16 ni -24 ga qo'shish.

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

-8 ning kvadrat ildizini chiqarish.

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

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{-4±2\sqrt{2}i}{-4}$$

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

$$x=\frac{-4+2\sqrt{2}i}{-4}$$

-4+2i\sqrt{2} ni -4 ga bo'lish.

$$x=-\frac{\sqrt{2}i}{2}+1$$

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

$$x=\frac{-2\sqrt{2}i-4}{-4}$$

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

$$x=\frac{\sqrt{2}i}{2}+1$$

Tenglama yechildi.

$$x=-\frac{\sqrt{2}i}{2}+1 x=\frac{\sqrt{2}i}{2}+1$$

Kvadratni yakunlash bo'yicha qadamlar

Tenglamaning ikkala tarafini 2 ga ko'paytirish.

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2x-1\right)^{2} kengaytirilishi uchun ishlating.

$$2x^{2}-\left(4x^{2}-4x+1\right)=2$$

4x^{2}-4x+1 teskarisini topish uchun har birining teskarisini toping.

$$2x^{2}-4x^{2}+4x-1=2$$

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

$$-2x^{2}+4x-1=2$$

1 ni ikki tarafga qo’shing.

$$-2x^{2}+4x=2+1$$

3 olish uchun 2 va 1'ni qo'shing.

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

Ikki tarafini -2 ga bo‘ling.

$$\frac{-2x^{2}+4x}{-2}=\frac{3}{-2}$$

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

$$x^{2}+\frac{4}{-2}x=\frac{3}{-2}$$

4 ni -2 ga bo'lish.

$$x^{2}-2x=\frac{3}{-2}$$

3 ni -2 ga bo'lish.

$$x^{2}-2x=-\frac{3}{2}$$

-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.

$$x^{2}-2x+1=-\frac{3}{2}+1$$

-\frac{3}{2} ni 1 ga qo'shish.

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

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-1=\frac{\sqrt{2}i}{2} x-1=-\frac{\sqrt{2}i}{2}$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{2}i}{2}+1 x=-\frac{\sqrt{2}i}{2}+1$$