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

$$( x - 2 ) ^ { 2 } = 1 + x$$

x uchun yechish

$$x = \frac{\sqrt{13} + 5}{2} \approx 4,302775638$$

$$x=\frac{5-\sqrt{13}}{2}\approx 0,697224362$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

Ikkala tarafdan 1 ni ayirish.

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

3 olish uchun 4 dan 1 ni ayirish.

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

Ikkala tarafdan x ni ayirish.

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

-5x ni olish uchun -4x va -x ni birlashtirish.

$$x^{2}-5x+3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -5 ni b va 3 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 3}}{2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3}}{2}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-12}}{2}$$

25 ni -12 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{13}}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{13}}{2}$$

x=\frac{5±\sqrt{13}}{2} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{13} ga qo'shish.

$$x=\frac{\sqrt{13}+5}{2}$$

x=\frac{5±\sqrt{13}}{2} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{13} ni ayirish.

$$x=\frac{5-\sqrt{13}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

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

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

Ikkala tarafdan x ni ayirish.

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

-5x ni olish uchun -4x va -x ni birlashtirish.

$$x^{2}-5x+4=1$$

Ikkala tarafdan 4 ni ayirish.

$$x^{2}-5x=1-4$$

-3 olish uchun 1 dan 4 ni ayirish.

$$x^{2}-5x=-3$$

-5 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{2} olish uchun. Keyin, -\frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-3+\left(-\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{2} kvadratini chiqarish.

$$x^{2}-5x+\frac{25}{4}=-3+\frac{25}{4}$$

-3 ni \frac{25}{4} ga qo'shish.

$$x^{2}-5x+\frac{25}{4}=\frac{13}{4}$$

x^{2}-5x+\frac{25}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{13}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Qisqartirish.

$$x-\frac{5}{2}=\frac{\sqrt{13}}{2} x-\frac{5}{2}=-\frac{\sqrt{13}}{2}$$

\frac{5}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}$$