x^2+6x+2=? eching

$$x ^ { 2 } + 6 x + 2 = ?$$

Omil

$$\left(x-\left(-\sqrt{7}-3\right)\right)\left(x-\left(\sqrt{7}-3\right)\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

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

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.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 2}}{2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-8}}{2}$$

36 ni -8 ga qo'shish.

$$x=\frac{-6±\sqrt{28}}{2}$$

28 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{7}}{2}$$

x=\frac{-6±2\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{7} ga qo'shish.

$$x=\frac{2\sqrt{7}-6}{2}$$

-6+2\sqrt{7} ni 2 ga bo'lish.

$$x=\sqrt{7}-3$$

x=\frac{-6±2\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{7} ni ayirish.

$$x=\frac{-2\sqrt{7}-6}{2}$$

-6-2\sqrt{7} ni 2 ga bo'lish.

$$x=-\sqrt{7}-3$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -3+\sqrt{7} ga va x_{2} uchun -3-\sqrt{7} ga bo‘ling.

$$x^{2}+6x+2=\left(x-\left(\sqrt{7}-3\right)\right)\left(x-\left(-\sqrt{7}-3\right)\right)$$

Baholash

$$x^{2}+6x+2$$