8x^2-12x-8 eching

$$8 x ^ { 2 } - 12 x - 8$$

Omil

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

Yechim qadamlari

4 omili.

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

Hisoblang: 2x^{2}-3x-2. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2x^{2}+ax+bx-2 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=2\left(-2\right)=-4$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -4-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-4 2,-2$$

Har bir juftlik yigʻindisini hisoblang.

$$1-4=-3 2-2=0$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=-4 b=1$$

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

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

2x^{2}-4x ichida 2x ni ajrating.

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

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

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

Toʻliq ajratilgan ifodani qaytadan yozing.

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

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.

$$8x^{2}-12x-8=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 8\left(-8\right)}}{2\times 8}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 8\left(-8\right)}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144-32\left(-8\right)}}{2\times 8}$$

-32 ni -8 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+256}}{2\times 8}$$

144 ni 256 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{400}}{2\times 8}$$

400 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±20}{2\times 8}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±20}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{12±20}{16}$$

x=\frac{12±20}{16} tenglamasini yeching, bunda ± musbat. 12 ni 20 ga qo'shish.

$$x=\frac{32}{16}$$

32 ni 16 ga bo'lish.

$$x=2$$

x=\frac{12±20}{16} tenglamasini yeching, bunda ± manfiy. 12 dan 20 ni ayirish.

$$x=-\frac{8}{16}$$

\frac{-8}{16} ulushini 8 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 2 ga va x_{2} uchun -\frac{1}{2} ga bo‘ling.

$$8x^{2}-12x-8=8\left(x-2\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$8x^{2}-12x-8=8\left(x-2\right)\left(x+\frac{1}{2}\right)$$

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.

$$8x^{2}-12x-8=8\left(x-2\right)\times \frac{2x+1}{2}$$

8 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

$$8x^{2}-12x-8=4\left(x-2\right)\left(2x+1\right)$$

Baholash

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