14t^2+21t eching

$$14 t ^ { 2 } + 21 t$$

Omil

$$7t\left(2t+3\right)$$

Yechim qadamlari

7 omili.

$$7\left(2t^{2}+3t\right)$$

Hisoblang: 2t^{2}+3t. t omili.

$$t\left(2t+3\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$7t\left(2t+3\right)$$

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.

$$14t^{2}+21t=0$$

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.

$$t=\frac{-21±\sqrt{21^{2}}}{2\times 14}$$

21^{2} ning kvadrat ildizini chiqarish.

$$t=\frac{-21±21}{2\times 14}$$

2 ni 14 marotabaga ko'paytirish.

$$t=\frac{-21±21}{28}$$

t=\frac{-21±21}{28} tenglamasini yeching, bunda ± musbat. -21 ni 21 ga qo'shish.

$$t=\frac{0}{28}$$

0 ni 28 ga bo'lish.

$$t=0$$

t=\frac{-21±21}{28} tenglamasini yeching, bunda ± manfiy. -21 dan 21 ni ayirish.

$$t=-\frac{42}{28}$$

\frac{-42}{28} ulushini 14 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$t=-\frac{3}{2}$$

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{3}{2} ga bo‘ling.

$$14t^{2}+21t=14t\left(t-\left(-\frac{3}{2}\right)\right)$$

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

$$14t^{2}+21t=14t\left(t+\frac{3}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{2} ni t ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$14t^{2}+21t=14t\times \frac{2t+3}{2}$$

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

$$14t^{2}+21t=7t\left(2t+3\right)$$

Baholash

$$7t\left(2t+3\right)$$