4t^2-13t-12 eching

$$4 t ^ { 2 } - 13 t - 12$$

Omil

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

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 4t^{2}+at+bt-12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-13 ab=4\left(-12\right)=-48$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -48-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Har bir juftlik yigʻindisini hisoblang.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

Yechim – -13 yigʻindisini beruvchi juftlik.

$$a=-16 b=3$$

4t^{2}-13t-12 ni \left(4t^{2}-16t\right)+\left(3t-12\right) sifatida qaytadan yozish.

$$\left(4t^{2}-16t\right)+\left(3t-12\right)$$

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

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

Distributiv funktsiyasidan foydalangan holda t-4 umumiy terminini chiqaring.

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

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.

$$4t^{2}-13t-12=0$$

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.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

-13 kvadratini chiqarish.

$$t=\frac{-\left(-13\right)±\sqrt{169-4\times 4\left(-12\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$t=\frac{-\left(-13\right)±\sqrt{169-16\left(-12\right)}}{2\times 4}$$

-16 ni -12 marotabaga ko'paytirish.

$$t=\frac{-\left(-13\right)±\sqrt{169+192}}{2\times 4}$$

169 ni 192 ga qo'shish.

$$t=\frac{-\left(-13\right)±\sqrt{361}}{2\times 4}$$

361 ning kvadrat ildizini chiqarish.

$$t=\frac{-\left(-13\right)±19}{2\times 4}$$

-13 ning teskarisi 13 ga teng.

$$t=\frac{13±19}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$t=\frac{13±19}{8}$$

t=\frac{13±19}{8} tenglamasini yeching, bunda ± musbat. 13 ni 19 ga qo'shish.

$$t=\frac{32}{8}$$

32 ni 8 ga bo'lish.

$$t=4$$

t=\frac{13±19}{8} tenglamasini yeching, bunda ± manfiy. 13 dan 19 ni ayirish.

$$t=-\frac{6}{8}$$

\frac{-6}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

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

$$4t^{2}-13t-12=4\left(t-4\right)\left(t-\left(-\frac{3}{4}\right)\right)$$

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

$$4t^{2}-13t-12=4\left(t-4\right)\left(t+\frac{3}{4}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{4} ni t ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$4t^{2}-13t-12=4\left(t-4\right)\times \frac{4t+3}{4}$$

4 va 4 ichida eng katta umumiy 4 faktorini bekor qiling.

$$4t^{2}-13t-12=\left(t-4\right)\left(4t+3\right)$$

Baholash

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