6x^2-23x-4 eching

$$6 { x }^{ 2 } -23x-4$$

Omil

$$\left(x-4\right)\left(6x+1\right)$$

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.

$$a+b=-23 ab=6\left(-4\right)=-24$$

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 – -23 yigʻindisini beruvchi juftlik.

$$a=-24 b=1$$

6x^{2}-23x-4 ni \left(6x^{2}-24x\right)+\left(x-4\right) sifatida qaytadan yozish.

$$\left(6x^{2}-24x\right)+\left(x-4\right)$$

6x^{2}-24x ichida 6x ni ajrating.

$$6x\left(x-4\right)+x-4$$

Distributiv funktsiyasidan foydalangan holda x-4 umumiy terminini chiqaring.

$$\left(x-4\right)\left(6x+1\right)$$

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.

$$6x^{2}-23x-4=0$$

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.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

-23 kvadratini chiqarish.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 6\left(-4\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-23\right)±\sqrt{529-24\left(-4\right)}}{2\times 6}$$

-24 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-23\right)±\sqrt{529+96}}{2\times 6}$$

529 ni 96 ga qo'shish.

$$x=\frac{-\left(-23\right)±\sqrt{625}}{2\times 6}$$

625 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-23\right)±25}{2\times 6}$$

-23 ning teskarisi 23 ga teng.

$$x=\frac{23±25}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{23±25}{12}$$

x=\frac{23±25}{12} tenglamasini yeching, bunda ± musbat. 23 ni 25 ga qo'shish.

$$x=\frac{48}{12}$$

48 ni 12 ga bo'lish.

$$x=4$$

x=\frac{23±25}{12} tenglamasini yeching, bunda ± manfiy. 23 dan 25 ni ayirish.

$$x=-\frac{2}{12}$$

\frac{-2}{12} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{6}$$

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{1}{6} ga bo‘ling.

$$6x^{2}-23x-4=6\left(x-4\right)\left(x-\left(-\frac{1}{6}\right)\right)$$

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

$$6x^{2}-23x-4=6\left(x-4\right)\left(x+\frac{1}{6}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{6} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6x^{2}-23x-4=6\left(x-4\right)\times \frac{6x+1}{6}$$

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

$$6x^{2}-23x-4=\left(x-4\right)\left(6x+1\right)$$

Baholash

$$\left(x-4\right)\left(6x+1\right)$$