(x^2+12x+8) eching

$$( x ^ { 2 } + 12 x + 8 )$$

Omil

$$\left(x-\left(-2\sqrt{7}-6\right)\right)\left(x-\left(2\sqrt{7}-6\right)\right)$$

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.

$$x^{2}+12x+8=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\times 8}}{2}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 8}}{2}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-32}}{2}$$

144 ni -32 ga qo'shish.

$$x=\frac{-12±\sqrt{112}}{2}$$

112 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±4\sqrt{7}}{2}$$

x=\frac{-12±4\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. -12 ni 4\sqrt{7} ga qo'shish.

$$x=\frac{4\sqrt{7}-12}{2}$$

-12+4\sqrt{7} ni 2 ga bo'lish.

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

x=\frac{-12±4\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. -12 dan 4\sqrt{7} ni ayirish.

$$x=\frac{-4\sqrt{7}-12}{2}$$

-12-4\sqrt{7} ni 2 ga bo'lish.

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -6+2\sqrt{7} ga va x_{2} uchun -6-2\sqrt{7} ga bo‘ling.

$$x^{2}+12x+8=\left(x-\left(2\sqrt{7}-6\right)\right)\left(x-\left(-2\sqrt{7}-6\right)\right)$$

Baholash

$$x^{2}+12x+8$$