2x^2-13x-24 eching

$$2 x ^ { 2 } - 13 x - 24$$

Omil

$$\left(x-8\right)\left(2x+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2x^{2}+ax+bx-24 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-13 ab=2\left(-24\right)=-48$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -48-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-48 2,-24 3,-16 4,-12 6,-8$$

Har bir juftlik yigʻindisini hisoblang.

$$1-48=-47 2-24=-22 3-16=-13 4-12=-8 6-8=-2$$

Yechim – -13 yigʻindisini beruvchi juftlik.

$$a=-16 b=3$$

2x^{2}-13x-24 ni \left(2x^{2}-16x\right)+\left(3x-24\right) sifatida qaytadan yozish.

$$\left(2x^{2}-16x\right)+\left(3x-24\right)$$

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

$$2x\left(x-8\right)+3\left(x-8\right)$$

Distributiv funktsiyasidan foydalangan holda x-8 umumiy terminini chiqaring.

$$\left(x-8\right)\left(2x+3\right)$$

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.

$$2x^{2}-13x-24=0$$

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.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

-13 kvadratini chiqarish.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 2\left(-24\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-13\right)±\sqrt{169-8\left(-24\right)}}{2\times 2}$$

-8 ni -24 marotabaga ko'paytirish.

$$x=\frac{-\left(-13\right)±\sqrt{169+192}}{2\times 2}$$

169 ni 192 ga qo'shish.

$$x=\frac{-\left(-13\right)±\sqrt{361}}{2\times 2}$$

361 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-13\right)±19}{2\times 2}$$

-13 ning teskarisi 13 ga teng.

$$x=\frac{13±19}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{13±19}{4}$$

x=\frac{13±19}{4} tenglamasini yeching, bunda ± musbat. 13 ni 19 ga qo'shish.

$$x=\frac{32}{4}$$

32 ni 4 ga bo'lish.

$$x=8$$

x=\frac{13±19}{4} tenglamasini yeching, bunda ± manfiy. 13 dan 19 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 8 ga va x_{2} uchun -\frac{3}{2} ga bo‘ling.

$$2x^{2}-13x-24=2\left(x-8\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$2x^{2}-13x-24=2\left(x-8\right)\left(x+\frac{3}{2}\right)$$

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.

$$2x^{2}-13x-24=2\left(x-8\right)\times \frac{2x+3}{2}$$

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

$$2x^{2}-13x-24=\left(x-8\right)\left(2x+3\right)$$

Baholash

$$\left(x-8\right)\left(2x+3\right)$$