12x^2-x-20 eching

$$12 x ^ { 2 } - x - 20$$

Omil

$$\left(3x-4\right)\left(4x+5\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 12x^{2}+ax+bx-20 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=12\left(-20\right)=-240$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -240-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-240 2,-120 3,-80 4,-60 5,-48 6,-40 8,-30 10,-24 12,-20 15,-16$$

Har bir juftlik yigʻindisini hisoblang.

$$1-240=-239 2-120=-118 3-80=-77 4-60=-56 5-48=-43 6-40=-34 8-30=-22 10-24=-14 12-20=-8 15-16=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=-16 b=15$$

12x^{2}-x-20 ni \left(12x^{2}-16x\right)+\left(15x-20\right) sifatida qaytadan yozish.

$$\left(12x^{2}-16x\right)+\left(15x-20\right)$$

Birinchi guruhda 4x ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$4x\left(3x-4\right)+5\left(3x-4\right)$$

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

$$\left(3x-4\right)\left(4x+5\right)$$

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.

$$12x^{2}-x-20=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 12\left(-20\right)}}{2\times 12}$$

-4 ni 12 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-48\left(-20\right)}}{2\times 12}$$

-48 ni -20 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+960}}{2\times 12}$$

1 ni 960 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{961}}{2\times 12}$$

961 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±31}{2\times 12}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±31}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$x=\frac{1±31}{24}$$

x=\frac{1±31}{24} tenglamasini yeching, bunda ± musbat. 1 ni 31 ga qo'shish.

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

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

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

x=\frac{1±31}{24} tenglamasini yeching, bunda ± manfiy. 1 dan 31 ni ayirish.

$$x=-\frac{30}{24}$$

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

$$x=-\frac{5}{4}$$

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{5}{4} ga bo‘ling.

$$12x^{2}-x-20=12\left(x-\frac{4}{3}\right)\left(x-\left(-\frac{5}{4}\right)\right)$$

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

$$12x^{2}-x-20=12\left(x-\frac{4}{3}\right)\left(x+\frac{5}{4}\right)$$

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

$$12x^{2}-x-20=12\times \frac{3x-4}{3}\left(x+\frac{5}{4}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{4} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$12x^{2}-x-20=12\times \frac{3x-4}{3}\times \frac{4x+5}{4}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{3x-4}{3} ni \frac{4x+5}{4} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$12x^{2}-x-20=12\times \frac{\left(3x-4\right)\left(4x+5\right)}{3\times 4}$$

3 ni 4 marotabaga ko'paytirish.

$$12x^{2}-x-20=12\times \frac{\left(3x-4\right)\left(4x+5\right)}{12}$$

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

$$12x^{2}-x-20=\left(3x-4\right)\left(4x+5\right)$$

Baholash

$$\left(3x-4\right)\left(4x+5\right)$$