6x^2+6x+1 eching

$$6 x ^ { 2 } + 6 x + 1$$

Omil

$$6\left(x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{2}\right)\right)\left(x-\left(\frac{\sqrt{3}}{6}-\frac{1}{2}\right)\right)$$

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.

$$6x^{2}+6x+1=0$$

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.

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

6 kvadratini chiqarish.

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

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-24}}{2\times 6}$$

36 ni -24 ga qo'shish.

$$x=\frac{-6±\sqrt{12}}{2\times 6}$$

12 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{3}}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{3}}{12}$$

x=\frac{-6±2\sqrt{3}}{12} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}-6}{12}$$

-6+2\sqrt{3} ni 12 ga bo'lish.

$$x=\frac{\sqrt{3}}{6}-\frac{1}{2}$$

x=\frac{-6±2\sqrt{3}}{12} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{3} ni ayirish.

$$x=\frac{-2\sqrt{3}-6}{12}$$

-6-2\sqrt{3} ni 12 ga bo'lish.

$$x=-\frac{\sqrt{3}}{6}-\frac{1}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{1}{2}+\frac{\sqrt{3}}{6} ga va x_{2} uchun -\frac{1}{2}-\frac{\sqrt{3}}{6} ga bo‘ling.

$$6x^{2}+6x+1=6\left(x-\left(\frac{\sqrt{3}}{6}-\frac{1}{2}\right)\right)\left(x-\left(-\frac{\sqrt{3}}{6}-\frac{1}{2}\right)\right)$$

Baholash

$$6x^{2}+6x+1$$