2x^2-5x-88 eching

$$2 { x }^{ 2 } -5x-88$$

Omil

$$\left(x-8\right)\left(2x+11\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2x^{2}+ax+bx-88 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=2\left(-88\right)=-176$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -176-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-176 2,-88 4,-44 8,-22 11,-16$$

Har bir juftlik yigʻindisini hisoblang.

$$1-176=-175 2-88=-86 4-44=-40 8-22=-14 11-16=-5$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=-16 b=11$$

2x^{2}-5x-88 ni \left(2x^{2}-16x\right)+\left(11x-88\right) sifatida qaytadan yozish.

$$\left(2x^{2}-16x\right)+\left(11x-88\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda 11 ni faktordan chiqaring.

$$2x\left(x-8\right)+11\left(x-8\right)$$

Distributiv funktsiyasidan foydalangan holda x-8 umumiy terminini chiqaring.

$$\left(x-8\right)\left(2x+11\right)$$

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.

$$2x^{2}-5x-88=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-88\right)}}{2\times 2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-88\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-88\right)}}{2\times 2}$$

-8 ni -88 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+704}}{2\times 2}$$

25 ni 704 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{729}}{2\times 2}$$

729 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±27}{2\times 2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±27}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{5±27}{4}$$

x=\frac{5±27}{4} tenglamasini yeching, bunda ± musbat. 5 ni 27 ga qo'shish.

$$x=\frac{32}{4}$$

32 ni 4 ga bo'lish.

$$x=8$$

x=\frac{5±27}{4} tenglamasini yeching, bunda ± manfiy. 5 dan 27 ni ayirish.

$$x=-\frac{22}{4}$$

\frac{-22}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{11}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 8 ga va x_{2} uchun -\frac{11}{2} ga bo‘ling.

$$2x^{2}-5x-88=2\left(x-8\right)\left(x-\left(-\frac{11}{2}\right)\right)$$

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

$$2x^{2}-5x-88=2\left(x-8\right)\left(x+\frac{11}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{11}{2} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$2x^{2}-5x-88=2\left(x-8\right)\times \frac{2x+11}{2}$$

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

$$2x^{2}-5x-88=\left(x-8\right)\left(2x+11\right)$$

Baholash

$$\left(x-8\right)\left(2x+11\right)$$