11x^2-20x-4 eching

$$11 x ^ { 2 } - 20 x - 4$$

Omil

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

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 11x^{2}+ax+bx-4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-20 ab=11\left(-4\right)=-44$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -44-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-44 2,-22 4,-11$$

Har bir juftlik yigʻindisini hisoblang.

$$1-44=-43 2-22=-20 4-11=-7$$

Yechim – -20 yigʻindisini beruvchi juftlik.

$$a=-22 b=2$$

11x^{2}-20x-4 ni \left(11x^{2}-22x\right)+\left(2x-4\right) sifatida qaytadan yozish.

$$\left(11x^{2}-22x\right)+\left(2x-4\right)$$

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

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

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

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

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.

$$11x^{2}-20x-4=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 11\left(-4\right)}}{2\times 11}$$

-20 kvadratini chiqarish.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 11\left(-4\right)}}{2\times 11}$$

-4 ni 11 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400-44\left(-4\right)}}{2\times 11}$$

-44 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400+176}}{2\times 11}$$

400 ni 176 ga qo'shish.

$$x=\frac{-\left(-20\right)±\sqrt{576}}{2\times 11}$$

576 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-20\right)±24}{2\times 11}$$

-20 ning teskarisi 20 ga teng.

$$x=\frac{20±24}{2\times 11}$$

2 ni 11 marotabaga ko'paytirish.

$$x=\frac{20±24}{22}$$

x=\frac{20±24}{22} tenglamasini yeching, bunda ± musbat. 20 ni 24 ga qo'shish.

$$x=\frac{44}{22}$$

44 ni 22 ga bo'lish.

$$x=2$$

x=\frac{20±24}{22} tenglamasini yeching, bunda ± manfiy. 20 dan 24 ni ayirish.

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

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

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

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{2}{11} ga bo‘ling.

$$11x^{2}-20x-4=11\left(x-2\right)\left(x-\left(-\frac{2}{11}\right)\right)$$

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

$$11x^{2}-20x-4=11\left(x-2\right)\left(x+\frac{2}{11}\right)$$

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

$$11x^{2}-20x-4=11\left(x-2\right)\times \frac{11x+2}{11}$$

11 va 11 ichida eng katta umumiy 11 faktorini bekor qiling.

$$11x^{2}-20x-4=\left(x-2\right)\left(11x+2\right)$$

Baholash

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