a^2+12a+4 eching

$$a ^ { 2 } + 12 a + 4$$

Omil

$$\left(a-\left(-4\sqrt{2}-6\right)\right)\left(a-\left(4\sqrt{2}-6\right)\right)$$

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.

$$a^{2}+12a+4=0$$

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.

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

12 kvadratini chiqarish.

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

-4 ni 4 marotabaga ko'paytirish.

$$a=\frac{-12±\sqrt{144-16}}{2}$$

144 ni -16 ga qo'shish.

$$a=\frac{-12±\sqrt{128}}{2}$$

128 ning kvadrat ildizini chiqarish.

$$a=\frac{-12±8\sqrt{2}}{2}$$

a=\frac{-12±8\sqrt{2}}{2} tenglamasini yeching, bunda ± musbat. -12 ni 8\sqrt{2} ga qo'shish.

$$a=\frac{8\sqrt{2}-12}{2}$$

-12+8\sqrt{2} ni 2 ga bo'lish.

$$a=4\sqrt{2}-6$$

a=\frac{-12±8\sqrt{2}}{2} tenglamasini yeching, bunda ± manfiy. -12 dan 8\sqrt{2} ni ayirish.

$$a=\frac{-8\sqrt{2}-12}{2}$$

-12-8\sqrt{2} ni 2 ga bo'lish.

$$a=-4\sqrt{2}-6$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -6+4\sqrt{2} ga va x_{2} uchun -6-4\sqrt{2} ga bo‘ling.

$$a^{2}+12a+4=\left(a-\left(4\sqrt{2}-6\right)\right)\left(a-\left(-4\sqrt{2}-6\right)\right)$$

Baholash

$$a^{2}+12a+4$$