x^2-4x+1=0 eching

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

x uchun yechish

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

$$x=2-\sqrt{3}\approx 0,267949192$$

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.

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

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

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

-4 kvadratini chiqarish.

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

16 ni -4 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{12}}{2}$$

12 ning kvadrat ildizini chiqarish.

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

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±2\sqrt{3}}{2}$$

x=\frac{4±2\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. 4 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}+4}{2}$$

4+2\sqrt{3} ni 2 ga bo'lish.

$$x=\sqrt{3}+2$$

x=\frac{4±2\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. 4 dan 2\sqrt{3} ni ayirish.

$$x=\frac{4-2\sqrt{3}}{2}$$

4-2\sqrt{3} ni 2 ga bo'lish.

$$x=2-\sqrt{3}$$

Tenglama yechildi.

$$x=\sqrt{3}+2 x=2-\sqrt{3}$$

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.

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

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

O‘zidan 1 ayirilsa 0 qoladi.

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

-4 ni bo‘lish, x shartining koeffitsienti, 2 ga -2 olish uchun. Keyin, -2 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

-2 kvadratini chiqarish.

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

-1 ni 4 ga qo'shish.

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

x^{2}-4x+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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-2=\sqrt{3} x-2=-\sqrt{3}$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{3}+2 x=2-\sqrt{3}$$