2x^2-11x-21 eching

$$2 x ^ { 2 } - 11 x - 21$$

Omil

$$\left(x-7\right)\left(2x+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2x^{2}+ax+bx-21 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=2\left(-21\right)=-42$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -42-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-42 2,-21 3,-14 6,-7$$

Har bir juftlik yigʻindisini hisoblang.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-14 b=3$$

2x^{2}-11x-21 ni \left(2x^{2}-14x\right)+\left(3x-21\right) sifatida qaytadan yozish.

$$\left(2x^{2}-14x\right)+\left(3x-21\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$2x\left(x-7\right)+3\left(x-7\right)$$

Distributiv funktsiyasidan foydalangan holda x-7 umumiy terminini chiqaring.

$$\left(x-7\right)\left(2x+3\right)$$

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.

$$2x^{2}-11x-21=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-21\right)}}{2\times 2}$$

-11 kvadratini chiqarish.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-21\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\left(-21\right)}}{2\times 2}$$

-8 ni -21 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121+168}}{2\times 2}$$

121 ni 168 ga qo'shish.

$$x=\frac{-\left(-11\right)±\sqrt{289}}{2\times 2}$$

289 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-11\right)±17}{2\times 2}$$

-11 ning teskarisi 11 ga teng.

$$x=\frac{11±17}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{11±17}{4}$$

x=\frac{11±17}{4} tenglamasini yeching, bunda ± musbat. 11 ni 17 ga qo'shish.

$$x=\frac{28}{4}$$

28 ni 4 ga bo'lish.

$$x=7$$

x=\frac{11±17}{4} tenglamasini yeching, bunda ± manfiy. 11 dan 17 ni ayirish.

$$x=-\frac{6}{4}$$

\frac{-6}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{3}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 7 ga va x_{2} uchun -\frac{3}{2} ga bo‘ling.

$$2x^{2}-11x-21=2\left(x-7\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$2x^{2}-11x-21=2\left(x-7\right)\left(x+\frac{3}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{2} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$2x^{2}-11x-21=2\left(x-7\right)\times \frac{2x+3}{2}$$

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

$$2x^{2}-11x-21=\left(x-7\right)\left(2x+3\right)$$

Baholash

$$\left(x-7\right)\left(2x+3\right)$$