5x^2-41x+42 eching

$$5 { x }^{ 2 } -41x+42$$

Omil

$$\left(x-7\right)\left(5x-6\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 5x^{2}+ax+bx+42 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-41 ab=5\times 42=210$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 210-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-210 -2,-105 -3,-70 -5,-42 -6,-35 -7,-30 -10,-21 -14,-15$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-210=-211 -2-105=-107 -3-70=-73 -5-42=-47 -6-35=-41 -7-30=-37 -10-21=-31 -14-15=-29$$

Yechim – -41 yigʻindisini beruvchi juftlik.

$$a=-35 b=-6$$

5x^{2}-41x+42 ni \left(5x^{2}-35x\right)+\left(-6x+42\right) sifatida qaytadan yozish.

$$\left(5x^{2}-35x\right)+\left(-6x+42\right)$$

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

$$5x\left(x-7\right)-6\left(x-7\right)$$

Distributiv funktsiyasidan foydalangan holda x-7 umumiy terminini chiqaring.

$$\left(x-7\right)\left(5x-6\right)$$

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.

$$5x^{2}-41x+42=0$$

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.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 5\times 42}}{2\times 5}$$

-41 kvadratini chiqarish.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 5\times 42}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-41\right)±\sqrt{1681-20\times 42}}{2\times 5}$$

-20 ni 42 marotabaga ko'paytirish.

$$x=\frac{-\left(-41\right)±\sqrt{1681-840}}{2\times 5}$$

1681 ni -840 ga qo'shish.

$$x=\frac{-\left(-41\right)±\sqrt{841}}{2\times 5}$$

841 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-41\right)±29}{2\times 5}$$

-41 ning teskarisi 41 ga teng.

$$x=\frac{41±29}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{41±29}{10}$$

x=\frac{41±29}{10} tenglamasini yeching, bunda ± musbat. 41 ni 29 ga qo'shish.

$$x=\frac{70}{10}$$

70 ni 10 ga bo'lish.

$$x=7$$

x=\frac{41±29}{10} tenglamasini yeching, bunda ± manfiy. 41 dan 29 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 7 ga va x_{2} uchun \frac{6}{5} ga bo‘ling.

$$5x^{2}-41x+42=5\left(x-7\right)\left(x-\frac{6}{5}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{6}{5} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$5x^{2}-41x+42=5\left(x-7\right)\times \frac{5x-6}{5}$$

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

$$5x^{2}-41x+42=\left(x-7\right)\left(5x-6\right)$$

Baholash

$$\left(x-7\right)\left(5x-6\right)$$