3x^2+8x+2 eching

$$3 { x }^{ 2 } +8x+2$$

Omil

$$3\left(x-\frac{-\sqrt{10}-4}{3}\right)\left(x-\frac{\sqrt{10}-4}{3}\right)$$

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.

$$3x^{2}+8x+2=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\times 2}}{2\times 3}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 3\times 2}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-12\times 2}}{2\times 3}$$

-12 ni 2 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-24}}{2\times 3}$$

64 ni -24 ga qo'shish.

$$x=\frac{-8±\sqrt{40}}{2\times 3}$$

40 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{10}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-8±2\sqrt{10}}{6}$$

x=\frac{-8±2\sqrt{10}}{6} tenglamasini yeching, bunda ± musbat. -8 ni 2\sqrt{10} ga qo'shish.

$$x=\frac{2\sqrt{10}-8}{6}$$

-8+2\sqrt{10} ni 6 ga bo'lish.

$$x=\frac{\sqrt{10}-4}{3}$$

x=\frac{-8±2\sqrt{10}}{6} tenglamasini yeching, bunda ± manfiy. -8 dan 2\sqrt{10} ni ayirish.

$$x=\frac{-2\sqrt{10}-8}{6}$$

-8-2\sqrt{10} ni 6 ga bo'lish.

$$x=\frac{-\sqrt{10}-4}{3}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-4+\sqrt{10}}{3} ga va x_{2} uchun \frac{-4-\sqrt{10}}{3} ga bo‘ling.

$$3x^{2}+8x+2=3\left(x-\frac{\sqrt{10}-4}{3}\right)\left(x-\frac{-\sqrt{10}-4}{3}\right)$$

Baholash

$$3x^{2}+8x+2$$