10x^2+20x eching

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

Omil

$$10x\left(x+2\right)$$

Yechim qadamlari

10 omili.

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

Hisoblang: x^{2}+2x. x omili.

$$x\left(x+2\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$10x\left(x+2\right)$$

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.

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

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.

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

20^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-20±20}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$x=\frac{-20±20}{20}$$

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

$$x=\frac{0}{20}$$

0 ni 20 ga bo'lish.

$$x=0$$

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

$$x=-\frac{40}{20}$$

-40 ni 20 ga bo'lish.

$$x=-2$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 0 ga va x_{2} uchun -2 ga bo‘ling.

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

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

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

Baholash

$$10x\left(x+2\right)$$