2x^2-8x+4 eching

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

Omil

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

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.

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

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 4}}{2\times 2}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 4}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\times 4}}{2\times 2}$$

-8 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64-32}}{2\times 2}$$

64 ni -32 ga qo'shish.

$$x=\frac{-\left(-8\right)±\sqrt{32}}{2\times 2}$$

32 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-8\right)±4\sqrt{2}}{2\times 2}$$

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±4\sqrt{2}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{8±4\sqrt{2}}{4}$$

x=\frac{8±4\sqrt{2}}{4} tenglamasini yeching, bunda ± musbat. 8 ni 4\sqrt{2} ga qo'shish.

$$x=\frac{4\sqrt{2}+8}{4}$$

8+4\sqrt{2} ni 4 ga bo'lish.

$$x=\sqrt{2}+2$$

x=\frac{8±4\sqrt{2}}{4} tenglamasini yeching, bunda ± manfiy. 8 dan 4\sqrt{2} ni ayirish.

$$x=\frac{8-4\sqrt{2}}{4}$$

8-4\sqrt{2} ni 4 ga bo'lish.

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

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

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

Baholash

$$2\left(x^{2}-4x+2\right)$$