n^2+8n eching

$$n ^ { 2 } + 8 n$$

Omil

$$n\left(n+8\right)$$

Yechim qadamlari

n omili.

$$n\left(n+8\right)$$

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.

$$n^{2}+8n=0$$

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.

$$n=\frac{-8±\sqrt{8^{2}}}{2}$$

8^{2} ning kvadrat ildizini chiqarish.

$$n=\frac{-8±8}{2}$$

n=\frac{-8±8}{2} tenglamasini yeching, bunda ± musbat. -8 ni 8 ga qo'shish.

$$n=\frac{0}{2}$$

0 ni 2 ga bo'lish.

$$n=0$$

n=\frac{-8±8}{2} tenglamasini yeching, bunda ± manfiy. -8 dan 8 ni ayirish.

$$n=-\frac{16}{2}$$

-16 ni 2 ga bo'lish.

$$n=-8$$

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 -8 ga bo‘ling.

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

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

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

Baholash

$$n\left(n+8\right)$$