12z^2-7z-12 eching

$$12 { z }^{ 2 } -7z-12$$

Omil

$$\left(3z-4\right)\left(4z+3\right)$$

Yechim qadamlari

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

$$a+b=-7 ab=12\left(-12\right)=-144$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -144-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-144 2,-72 3,-48 4,-36 6,-24 8,-18 9,-16 12,-12$$

Har bir juftlik yigʻindisini hisoblang.

$$1-144=-143 2-72=-70 3-48=-45 4-36=-32 6-24=-18 8-18=-10 9-16=-7 12-12=0$$

Yechim – -7 yigʻindisini beruvchi juftlik.

$$a=-16 b=9$$

12z^{2}-7z-12 ni \left(12z^{2}-16z\right)+\left(9z-12\right) sifatida qaytadan yozish.

$$\left(12z^{2}-16z\right)+\left(9z-12\right)$$

Birinchi guruhda 4z ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$4z\left(3z-4\right)+3\left(3z-4\right)$$

Distributiv funktsiyasidan foydalangan holda 3z-4 umumiy terminini chiqaring.

$$\left(3z-4\right)\left(4z+3\right)$$

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.

$$12z^{2}-7z-12=0$$

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.

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

-7 kvadratini chiqarish.

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

-4 ni 12 marotabaga ko'paytirish.

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

-48 ni -12 marotabaga ko'paytirish.

$$z=\frac{-\left(-7\right)±\sqrt{49+576}}{2\times 12}$$

49 ni 576 ga qo'shish.

$$z=\frac{-\left(-7\right)±\sqrt{625}}{2\times 12}$$

625 ning kvadrat ildizini chiqarish.

$$z=\frac{-\left(-7\right)±25}{2\times 12}$$

-7 ning teskarisi 7 ga teng.

$$z=\frac{7±25}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$z=\frac{7±25}{24}$$

z=\frac{7±25}{24} tenglamasini yeching, bunda ± musbat. 7 ni 25 ga qo'shish.

$$z=\frac{32}{24}$$

\frac{32}{24} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$z=\frac{4}{3}$$

z=\frac{7±25}{24} tenglamasini yeching, bunda ± manfiy. 7 dan 25 ni ayirish.

$$z=-\frac{18}{24}$$

\frac{-18}{24} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$z=-\frac{3}{4}$$

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

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

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

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

Umumiy maxrajni topib va suratlarni ayirib \frac{4}{3} ni z dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{4} ni z ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$12z^{2}-7z-12=12\times \frac{3z-4}{3}\times \frac{4z+3}{4}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{3z-4}{3} ni \frac{4z+3}{4} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

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

3 ni 4 marotabaga ko'paytirish.

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

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

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

Baholash

$$\left(3z-4\right)\left(4z+3\right)$$