8x^2+16x+4 eching

$$8 x ^ { 2 } + 16 x + 4$$

Omil

$$8\left(x-\left(-\frac{\sqrt{2}}{2}-1\right)\right)\left(x-\left(\frac{\sqrt{2}}{2}-1\right)\right)$$

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.

$$8x^{2}+16x+4=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 8\times 4}}{2\times 8}$$

16 kvadratini chiqarish.

$$x=\frac{-16±\sqrt{256-4\times 8\times 4}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-16±\sqrt{256-32\times 4}}{2\times 8}$$

-32 ni 4 marotabaga ko'paytirish.

$$x=\frac{-16±\sqrt{256-128}}{2\times 8}$$

256 ni -128 ga qo'shish.

$$x=\frac{-16±\sqrt{128}}{2\times 8}$$

128 ning kvadrat ildizini chiqarish.

$$x=\frac{-16±8\sqrt{2}}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{-16±8\sqrt{2}}{16}$$

x=\frac{-16±8\sqrt{2}}{16} tenglamasini yeching, bunda ± musbat. -16 ni 8\sqrt{2} ga qo'shish.

$$x=\frac{8\sqrt{2}-16}{16}$$

-16+8\sqrt{2} ni 16 ga bo'lish.

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

x=\frac{-16±8\sqrt{2}}{16} tenglamasini yeching, bunda ± manfiy. -16 dan 8\sqrt{2} ni ayirish.

$$x=\frac{-8\sqrt{2}-16}{16}$$

-16-8\sqrt{2} ni 16 ga bo'lish.

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

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

$$8x^{2}+16x+4=8\left(x-\left(\frac{\sqrt{2}}{2}-1\right)\right)\left(x-\left(-\frac{\sqrt{2}}{2}-1\right)\right)$$

Baholash

$$8x^{2}+16x+4$$