11x^2+140x-196= eching

$$11 { x }^{ 2 } +140x-196=$$

Omil

$$\left(11x-14\right)\left(x+14\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 11x^{2}+ax+bx-196 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=140 ab=11\left(-196\right)=-2156$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -2156-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,2156 -2,1078 -4,539 -7,308 -11,196 -14,154 -22,98 -28,77 -44,49$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+2156=2155 -2+1078=1076 -4+539=535 -7+308=301 -11+196=185 -14+154=140 -22+98=76 -28+77=49 -44+49=5$$

Yechim – 140 yigʻindisini beruvchi juftlik.

$$a=-14 b=154$$

11x^{2}+140x-196 ni \left(11x^{2}-14x\right)+\left(154x-196\right) sifatida qaytadan yozish.

$$\left(11x^{2}-14x\right)+\left(154x-196\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 14 ni faktordan chiqaring.

$$x\left(11x-14\right)+14\left(11x-14\right)$$

Distributiv funktsiyasidan foydalangan holda 11x-14 umumiy terminini chiqaring.

$$\left(11x-14\right)\left(x+14\right)$$

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.

$$11x^{2}+140x-196=0$$

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.

$$x=\frac{-140±\sqrt{140^{2}-4\times 11\left(-196\right)}}{2\times 11}$$

140 kvadratini chiqarish.

$$x=\frac{-140±\sqrt{19600-4\times 11\left(-196\right)}}{2\times 11}$$

-4 ni 11 marotabaga ko'paytirish.

$$x=\frac{-140±\sqrt{19600-44\left(-196\right)}}{2\times 11}$$

-44 ni -196 marotabaga ko'paytirish.

$$x=\frac{-140±\sqrt{19600+8624}}{2\times 11}$$

19600 ni 8624 ga qo'shish.

$$x=\frac{-140±\sqrt{28224}}{2\times 11}$$

28224 ning kvadrat ildizini chiqarish.

$$x=\frac{-140±168}{2\times 11}$$

2 ni 11 marotabaga ko'paytirish.

$$x=\frac{-140±168}{22}$$

x=\frac{-140±168}{22} tenglamasini yeching, bunda ± musbat. -140 ni 168 ga qo'shish.

$$x=\frac{28}{22}$$

\frac{28}{22} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{14}{11}$$

x=\frac{-140±168}{22} tenglamasini yeching, bunda ± manfiy. -140 dan 168 ni ayirish.

$$x=-\frac{308}{22}$$

-308 ni 22 ga bo'lish.

$$x=-14$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{14}{11} ga va x_{2} uchun -14 ga bo‘ling.

$$11x^{2}+140x-196=11\left(x-\frac{14}{11}\right)\left(x-\left(-14\right)\right)$$

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

$$11x^{2}+140x-196=11\left(x-\frac{14}{11}\right)\left(x+14\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{14}{11} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$11x^{2}+140x-196=11\times \frac{11x-14}{11}\left(x+14\right)$$

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

$$11x^{2}+140x-196=\left(11x-14\right)\left(x+14\right)$$

Baholash

$$\left(11x-14\right)\left(x+14\right)$$