10n^2+25n eching

$$10 n ^ { 2 } + 25 n$$

Omil

$$5n\left(2n+5\right)$$

Yechim qadamlari

5 omili.

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

Hisoblang: 2n^{2}+5n. n omili.

$$n\left(2n+5\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$5n\left(2n+5\right)$$

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.

$$10n^{2}+25n=0$$

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.

$$n=\frac{-25±\sqrt{25^{2}}}{2\times 10}$$

25^{2} ning kvadrat ildizini chiqarish.

$$n=\frac{-25±25}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$n=\frac{-25±25}{20}$$

n=\frac{-25±25}{20} tenglamasini yeching, bunda ± musbat. -25 ni 25 ga qo'shish.

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

0 ni 20 ga bo'lish.

$$n=0$$

n=\frac{-25±25}{20} tenglamasini yeching, bunda ± manfiy. -25 dan 25 ni ayirish.

$$n=-\frac{50}{20}$$

\frac{-50}{20} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=-\frac{5}{2}$$

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 -\frac{5}{2} ga bo‘ling.

$$10n^{2}+25n=10n\left(n-\left(-\frac{5}{2}\right)\right)$$

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

$$10n^{2}+25n=10n\left(n+\frac{5}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni n ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$10n^{2}+25n=10n\times \frac{2n+5}{2}$$

10 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

$$10n^{2}+25n=5n\left(2n+5\right)$$

Baholash

$$5n\left(2n+5\right)$$