3q^2-143q+1602 eching

$$3 q ^ { 2 } - 143 q + 1602$$

Omil

$$\left(q-18\right)\left(3q-89\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3q^{2}+aq+bq+1602 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-143 ab=3\times 1602=4806$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 4806-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-4806 -2,-2403 -3,-1602 -6,-801 -9,-534 -18,-267 -27,-178 -54,-89$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-4806=-4807 -2-2403=-2405 -3-1602=-1605 -6-801=-807 -9-534=-543 -18-267=-285 -27-178=-205 -54-89=-143$$

Yechim – -143 yigʻindisini beruvchi juftlik.

$$a=-89 b=-54$$

3q^{2}-143q+1602 ni \left(3q^{2}-89q\right)+\left(-54q+1602\right) sifatida qaytadan yozish.

$$\left(3q^{2}-89q\right)+\left(-54q+1602\right)$$

Birinchi guruhda q ni va ikkinchi guruhda -18 ni faktordan chiqaring.

$$q\left(3q-89\right)-18\left(3q-89\right)$$

Distributiv funktsiyasidan foydalangan holda 3q-89 umumiy terminini chiqaring.

$$\left(3q-89\right)\left(q-18\right)$$

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.

$$3q^{2}-143q+1602=0$$

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.

$$q=\frac{-\left(-143\right)±\sqrt{\left(-143\right)^{2}-4\times 3\times 1602}}{2\times 3}$$

-143 kvadratini chiqarish.

$$q=\frac{-\left(-143\right)±\sqrt{20449-4\times 3\times 1602}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$q=\frac{-\left(-143\right)±\sqrt{20449-12\times 1602}}{2\times 3}$$

-12 ni 1602 marotabaga ko'paytirish.

$$q=\frac{-\left(-143\right)±\sqrt{20449-19224}}{2\times 3}$$

20449 ni -19224 ga qo'shish.

$$q=\frac{-\left(-143\right)±\sqrt{1225}}{2\times 3}$$

1225 ning kvadrat ildizini chiqarish.

$$q=\frac{-\left(-143\right)±35}{2\times 3}$$

-143 ning teskarisi 143 ga teng.

$$q=\frac{143±35}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$q=\frac{143±35}{6}$$

q=\frac{143±35}{6} tenglamasini yeching, bunda ± musbat. 143 ni 35 ga qo'shish.

$$q=\frac{178}{6}$$

\frac{178}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$q=\frac{89}{3}$$

q=\frac{143±35}{6} tenglamasini yeching, bunda ± manfiy. 143 dan 35 ni ayirish.

$$q=\frac{108}{6}$$

108 ni 6 ga bo'lish.

$$q=18$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{89}{3} ga va x_{2} uchun 18 ga bo‘ling.

$$3q^{2}-143q+1602=3\left(q-\frac{89}{3}\right)\left(q-18\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{89}{3} ni q dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$3q^{2}-143q+1602=3\times \frac{3q-89}{3}\left(q-18\right)$$

3 va 3 ichida eng katta umumiy 3 faktorini bekor qiling.

$$3q^{2}-143q+1602=\left(3q-89\right)\left(q-18\right)$$

Baholash

$$\left(q-18\right)\left(3q-89\right)$$