5x^2+12x-44= eching

$$5 x ^ { 2 } + 12 x - 44 =$$

Omil

$$\left(x-2\right)\left(5x+22\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 5x^{2}+ax+bx-44 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=12 ab=5\left(-44\right)=-220$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -220-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,220 -2,110 -4,55 -5,44 -10,22 -11,20$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+220=219 -2+110=108 -4+55=51 -5+44=39 -10+22=12 -11+20=9$$

Yechim – 12 yigʻindisini beruvchi juftlik.

$$a=-10 b=22$$

5x^{2}+12x-44 ni \left(5x^{2}-10x\right)+\left(22x-44\right) sifatida qaytadan yozish.

$$\left(5x^{2}-10x\right)+\left(22x-44\right)$$

Birinchi guruhda 5x ni va ikkinchi guruhda 22 ni faktordan chiqaring.

$$5x\left(x-2\right)+22\left(x-2\right)$$

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

$$\left(x-2\right)\left(5x+22\right)$$

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.

$$5x^{2}+12x-44=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\left(-44\right)}}{2\times 5}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 5\left(-44\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-20\left(-44\right)}}{2\times 5}$$

-20 ni -44 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+880}}{2\times 5}$$

144 ni 880 ga qo'shish.

$$x=\frac{-12±\sqrt{1024}}{2\times 5}$$

1024 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±32}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-12±32}{10}$$

x=\frac{-12±32}{10} tenglamasini yeching, bunda ± musbat. -12 ni 32 ga qo'shish.

$$x=\frac{20}{10}$$

20 ni 10 ga bo'lish.

$$x=2$$

x=\frac{-12±32}{10} tenglamasini yeching, bunda ± manfiy. -12 dan 32 ni ayirish.

$$x=-\frac{44}{10}$$

\frac{-44}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{22}{5}$$

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{22}{5} ga bo‘ling.

$$5x^{2}+12x-44=5\left(x-2\right)\left(x-\left(-\frac{22}{5}\right)\right)$$

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

$$5x^{2}+12x-44=5\left(x-2\right)\left(x+\frac{22}{5}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{22}{5} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$5x^{2}+12x-44=5\left(x-2\right)\times \frac{5x+22}{5}$$

5 va 5 ichida eng katta umumiy 5 faktorini bekor qiling.

$$5x^{2}+12x-44=\left(x-2\right)\left(5x+22\right)$$

Baholash

$$\left(x-2\right)\left(5x+22\right)$$