6x^2-5x-4 eching

$$6 x ^ { 2 } - 5 x - 4$$

Omil

$$\left(3x-4\right)\left(2x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6x^{2}+ax+bx-4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=6\left(-4\right)=-24$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -24-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

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

Har bir juftlik yigʻindisini hisoblang.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=-8 b=3$$

6x^{2}-5x-4 ni \left(6x^{2}-8x\right)+\left(3x-4\right) sifatida qaytadan yozish.

$$\left(6x^{2}-8x\right)+\left(3x-4\right)$$

6x^{2}-8x ichida 2x ni ajrating.

$$2x\left(3x-4\right)+3x-4$$

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

$$\left(3x-4\right)\left(2x+1\right)$$

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.

$$6x^{2}-5x-4=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-4\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-24\left(-4\right)}}{2\times 6}$$

-24 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 6}$$

25 ni 96 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{121}}{2\times 6}$$

121 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±11}{2\times 6}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±11}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{5±11}{12}$$

x=\frac{5±11}{12} tenglamasini yeching, bunda ± musbat. 5 ni 11 ga qo'shish.

$$x=\frac{16}{12}$$

\frac{16}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{5±11}{12} tenglamasini yeching, bunda ± manfiy. 5 dan 11 ni ayirish.

$$x=-\frac{6}{12}$$

\frac{-6}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{2}$$

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{1}{2} ga bo‘ling.

$$6x^{2}-5x-4=6\left(x-\frac{4}{3}\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$6x^{2}-5x-4=6\left(x-\frac{4}{3}\right)\left(x+\frac{1}{2}\right)$$

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

$$6x^{2}-5x-4=6\times \frac{3x-4}{3}\left(x+\frac{1}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6x^{2}-5x-4=6\times \frac{3x-4}{3}\times \frac{2x+1}{2}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{3x-4}{3} ni \frac{2x+1}{2} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$6x^{2}-5x-4=6\times \frac{\left(3x-4\right)\left(2x+1\right)}{3\times 2}$$

3 ni 2 marotabaga ko'paytirish.

$$6x^{2}-5x-4=6\times \frac{\left(3x-4\right)\left(2x+1\right)}{6}$$

6 va 6 ichida eng katta umumiy 6 faktorini bekor qiling.

$$6x^{2}-5x-4=\left(3x-4\right)\left(2x+1\right)$$

Baholash

$$\left(3x-4\right)\left(2x+1\right)$$