4y^2+4y eching

$$4 y ^ { 2 } + 4 y$$

Omil

$$4y\left(y+1\right)$$

Yechim qadamlari

4 omili.

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

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

$$y\left(y+1\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$4y\left(y+1\right)$$

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.

$$4y^{2}+4y=0$$

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.

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

4^{2} ning kvadrat ildizini chiqarish.

$$y=\frac{-4±4}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$y=\frac{-4±4}{8}$$

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

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

0 ni 8 ga bo'lish.

$$y=0$$

y=\frac{-4±4}{8} tenglamasini yeching, bunda ± manfiy. -4 dan 4 ni ayirish.

$$y=-\frac{8}{8}$$

-8 ni 8 ga bo'lish.

$$y=-1$$

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 -1 ga bo‘ling.

$$4y^{2}+4y=4y\left(y-\left(-1\right)\right)$$

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

$$4y^{2}+4y=4y\left(y+1\right)$$

Baholash

$$4y\left(y+1\right)$$