12t^2-7t-10 eching

$$12 t ^ { 2 } - 7 t - 10$$

Omil

$$\left(4t-5\right)\left(3t+2\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 12t^{2}+at+bt-10 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-7 ab=12\left(-10\right)=-120$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -120-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12$$

Har bir juftlik yigʻindisini hisoblang.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

Yechim – -7 yigʻindisini beruvchi juftlik.

$$a=-15 b=8$$

12t^{2}-7t-10 ni \left(12t^{2}-15t\right)+\left(8t-10\right) sifatida qaytadan yozish.

$$\left(12t^{2}-15t\right)+\left(8t-10\right)$$

Birinchi guruhda 3t ni va ikkinchi guruhda 2 ni faktordan chiqaring.

$$3t\left(4t-5\right)+2\left(4t-5\right)$$

Distributiv funktsiyasidan foydalangan holda 4t-5 umumiy terminini chiqaring.

$$\left(4t-5\right)\left(3t+2\right)$$

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.

$$12t^{2}-7t-10=0$$

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.

$$t=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12\left(-10\right)}}{2\times 12}$$

-7 kvadratini chiqarish.

$$t=\frac{-\left(-7\right)±\sqrt{49-4\times 12\left(-10\right)}}{2\times 12}$$

-4 ni 12 marotabaga ko'paytirish.

$$t=\frac{-\left(-7\right)±\sqrt{49-48\left(-10\right)}}{2\times 12}$$

-48 ni -10 marotabaga ko'paytirish.

$$t=\frac{-\left(-7\right)±\sqrt{49+480}}{2\times 12}$$

49 ni 480 ga qo'shish.

$$t=\frac{-\left(-7\right)±\sqrt{529}}{2\times 12}$$

529 ning kvadrat ildizini chiqarish.

$$t=\frac{-\left(-7\right)±23}{2\times 12}$$

-7 ning teskarisi 7 ga teng.

$$t=\frac{7±23}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$t=\frac{7±23}{24}$$

t=\frac{7±23}{24} tenglamasini yeching, bunda ± musbat. 7 ni 23 ga qo'shish.

$$t=\frac{30}{24}$$

\frac{30}{24} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$t=\frac{5}{4}$$

t=\frac{7±23}{24} tenglamasini yeching, bunda ± manfiy. 7 dan 23 ni ayirish.

$$t=-\frac{16}{24}$$

\frac{-16}{24} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{5}{4} ga va x_{2} uchun -\frac{2}{3} ga bo‘ling.

$$12t^{2}-7t-10=12\left(t-\frac{5}{4}\right)\left(t-\left(-\frac{2}{3}\right)\right)$$

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

$$12t^{2}-7t-10=12\left(t-\frac{5}{4}\right)\left(t+\frac{2}{3}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{5}{4} ni t dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$12t^{2}-7t-10=12\times \frac{4t-5}{4}\left(t+\frac{2}{3}\right)$$

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

$$12t^{2}-7t-10=12\times \frac{4t-5}{4}\times \frac{3t+2}{3}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{4t-5}{4} ni \frac{3t+2}{3} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$12t^{2}-7t-10=12\times \frac{\left(4t-5\right)\left(3t+2\right)}{4\times 3}$$

4 ni 3 marotabaga ko'paytirish.

$$12t^{2}-7t-10=12\times \frac{\left(4t-5\right)\left(3t+2\right)}{12}$$

12 va 12 ichida eng katta umumiy 12 faktorini bekor qiling.

$$12t^{2}-7t-10=\left(4t-5\right)\left(3t+2\right)$$

Baholash

$$\left(4t-5\right)\left(3t+2\right)$$