4u^2+8u eching

$$4 u ^ { 2 } + 8 u$$

Omil

$$4u\left(u+2\right)$$

Yechim qadamlari

4 omili.

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

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

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

Toʻliq ajratilgan ifodani qaytadan yozing.

$$4u\left(u+2\right)$$

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.

$$4u^{2}+8u=0$$

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.

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

8^{2} ning kvadrat ildizini chiqarish.

$$u=\frac{-8±8}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$u=\frac{-8±8}{8}$$

u=\frac{-8±8}{8} tenglamasini yeching, bunda ± musbat. -8 ni 8 ga qo'shish.

$$u=\frac{0}{8}$$

0 ni 8 ga bo'lish.

$$u=0$$

u=\frac{-8±8}{8} tenglamasini yeching, bunda ± manfiy. -8 dan 8 ni ayirish.

$$u=-\frac{16}{8}$$

-16 ni 8 ga bo'lish.

$$u=-2$$

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.

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

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

$$4u^{2}+8u=4u\left(u+2\right)$$

Baholash

$$4u\left(u+2\right)$$