5x^2-4x-12 eching

$$5 x ^ { 2 } - 4 x - 12$$

Omil

$$\left(x-2\right)\left(5x+6\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 5x^{2}+ax+bx-12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-4 ab=5\left(-12\right)=-60$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -60-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

Yechim – -4 yigʻindisini beruvchi juftlik.

$$a=-10 b=6$$

5x^{2}-4x-12 ni \left(5x^{2}-10x\right)+\left(6x-12\right) sifatida qaytadan yozish.

$$\left(5x^{2}-10x\right)+\left(6x-12\right)$$

Birinchi guruhda 5x ni va ikkinchi guruhda 6 ni faktordan chiqaring.

$$5x\left(x-2\right)+6\left(x-2\right)$$

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

$$\left(x-2\right)\left(5x+6\right)$$

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.

$$5x^{2}-4x-12=0$$

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.

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

-4 kvadratini chiqarish.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\left(-12\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16-20\left(-12\right)}}{2\times 5}$$

-20 ni -12 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+240}}{2\times 5}$$

16 ni 240 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{256}}{2\times 5}$$

256 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-4\right)±16}{2\times 5}$$

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±16}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{4±16}{10}$$

x=\frac{4±16}{10} tenglamasini yeching, bunda ± musbat. 4 ni 16 ga qo'shish.

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

20 ni 10 ga bo'lish.

$$x=2$$

x=\frac{4±16}{10} tenglamasini yeching, bunda ± manfiy. 4 dan 16 ni ayirish.

$$x=-\frac{12}{10}$$

\frac{-12}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{6}{5}$$

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{6}{5} ga bo‘ling.

$$5x^{2}-4x-12=5\left(x-2\right)\left(x-\left(-\frac{6}{5}\right)\right)$$

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

$$5x^{2}-4x-12=5\left(x-2\right)\left(x+\frac{6}{5}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{6}{5} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$5x^{2}-4x-12=5\left(x-2\right)\times \frac{5x+6}{5}$$

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

$$5x^{2}-4x-12=\left(x-2\right)\left(5x+6\right)$$

Baholash

$$\left(x-2\right)\left(5x+6\right)$$