-2x^2-5x+7 eching

$$- 2 x ^ { 2 } - 5 x + 7$$

Omil

$$\left(1-x\right)\left(2x+7\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -2x^{2}+ax+bx+7 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=-2\times 7=-14$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -14-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-14 2,-7$$

Har bir juftlik yigʻindisini hisoblang.

$$1-14=-13 2-7=-5$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=2 b=-7$$

-2x^{2}-5x+7 ni \left(-2x^{2}+2x\right)+\left(-7x+7\right) sifatida qaytadan yozish.

$$\left(-2x^{2}+2x\right)+\left(-7x+7\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda 7 ni faktordan chiqaring.

$$2x\left(-x+1\right)+7\left(-x+1\right)$$

Distributiv funktsiyasidan foydalangan holda -x+1 umumiy terminini chiqaring.

$$\left(-x+1\right)\left(2x+7\right)$$

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.

$$-2x^{2}-5x+7=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\times 7}}{2\left(-2\right)}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 7}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\times 7}}{2\left(-2\right)}$$

8 ni 7 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+56}}{2\left(-2\right)}$$

25 ni 56 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{81}}{2\left(-2\right)}$$

81 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±9}{2\left(-2\right)}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±9}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{5±9}{-4}$$

x=\frac{5±9}{-4} tenglamasini yeching, bunda ± musbat. 5 ni 9 ga qo'shish.

$$x=\frac{14}{-4}$$

\frac{14}{-4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{7}{2}$$

x=\frac{5±9}{-4} tenglamasini yeching, bunda ± manfiy. 5 dan 9 ni ayirish.

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

-4 ni -4 ga bo'lish.

$$x=1$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{7}{2} ga va x_{2} uchun 1 ga bo‘ling.

$$-2x^{2}-5x+7=-2\left(x-\left(-\frac{7}{2}\right)\right)\left(x-1\right)$$

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

$$-2x^{2}-5x+7=-2\left(x+\frac{7}{2}\right)\left(x-1\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{7}{2} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$-2x^{2}-5x+7=-2\times \frac{-2x-7}{-2}\left(x-1\right)$$

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

$$-2x^{2}-5x+7=\left(-2x-7\right)\left(x-1\right)$$

Baholash

$$\left(1-x\right)\left(2x+7\right)$$