20x+2x^2+10 eching

$$20x+2 { x }^{ 2 } +10$$

Omil

$$2\left(x-\left(-2\sqrt{5}-5\right)\right)\left(x-\left(2\sqrt{5}-5\right)\right)$$

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.

$$2x^{2}+20x+10=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\times 10}}{2\times 2}$$

20 kvadratini chiqarish.

$$x=\frac{-20±\sqrt{400-4\times 2\times 10}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-20±\sqrt{400-8\times 10}}{2\times 2}$$

-8 ni 10 marotabaga ko'paytirish.

$$x=\frac{-20±\sqrt{400-80}}{2\times 2}$$

400 ni -80 ga qo'shish.

$$x=\frac{-20±\sqrt{320}}{2\times 2}$$

320 ning kvadrat ildizini chiqarish.

$$x=\frac{-20±8\sqrt{5}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-20±8\sqrt{5}}{4}$$

x=\frac{-20±8\sqrt{5}}{4} tenglamasini yeching, bunda ± musbat. -20 ni 8\sqrt{5} ga qo'shish.

$$x=\frac{8\sqrt{5}-20}{4}$$

-20+8\sqrt{5} ni 4 ga bo'lish.

$$x=2\sqrt{5}-5$$

x=\frac{-20±8\sqrt{5}}{4} tenglamasini yeching, bunda ± manfiy. -20 dan 8\sqrt{5} ni ayirish.

$$x=\frac{-8\sqrt{5}-20}{4}$$

-20-8\sqrt{5} ni 4 ga bo'lish.

$$x=-2\sqrt{5}-5$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -5+2\sqrt{5} ga va x_{2} uchun -5-2\sqrt{5} ga bo‘ling.

$$2x^{2}+20x+10=2\left(x-\left(2\sqrt{5}-5\right)\right)\left(x-\left(-2\sqrt{5}-5\right)\right)$$

Baholash

$$2\left(x^{2}+10x+5\right)$$