x^2-6x+4=0 eching

$$x ^ { 2 } - 6 x + 4 = 0$$

x uchun yechish

$$x=\sqrt{5}+3\approx 5,236067977$$

$$x=3-\sqrt{5}\approx 0,763932023$$

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.

$$x^{2}-6x+4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -6 ni b va 4 ni c bilan almashtiring.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4}}{2}$$

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 4}}{2}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36-16}}{2}$$

36 ni -16 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{20}}{2}$$

20 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-6\right)±2\sqrt{5}}{2}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±2\sqrt{5}}{2}$$

x=\frac{6±2\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. 6 ni 2\sqrt{5} ga qo'shish.

$$x=\frac{2\sqrt{5}+6}{2}$$

6+2\sqrt{5} ni 2 ga bo'lish.

$$x=\sqrt{5}+3$$

x=\frac{6±2\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. 6 dan 2\sqrt{5} ni ayirish.

$$x=\frac{6-2\sqrt{5}}{2}$$

6-2\sqrt{5} ni 2 ga bo'lish.

$$x=3-\sqrt{5}$$

Tenglama yechildi.

$$x=\sqrt{5}+3 x=3-\sqrt{5}$$

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.

$$x^{2}-6x+4=0$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$x^{2}-6x+4-4=-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$x^{2}-6x=-4$$

-6 ni bo‘lish, x shartining koeffitsienti, 2 ga -3 olish uchun. Keyin, -3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-6x+\left(-3\right)^{2}=-4+\left(-3\right)^{2}$$

-3 kvadratini chiqarish.

$$x^{2}-6x+9=-4+9$$

-4 ni 9 ga qo'shish.

$$x^{2}-6x+9=5$$

x^{2}-6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-3=\sqrt{5} x-3=-\sqrt{5}$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{5}+3 x=3-\sqrt{5}$$