20n^2-7n-3 eching

$$20 n ^ { 2 } - 7 n - 3$$

Omil

$$\left(5n-3\right)\left(4n+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 20n^{2}+an+bn-3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-7 ab=20\left(-3\right)=-60$$

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 – -7 yigʻindisini beruvchi juftlik.

$$a=-12 b=5$$

20n^{2}-7n-3 ni \left(20n^{2}-12n\right)+\left(5n-3\right) sifatida qaytadan yozish.

$$\left(20n^{2}-12n\right)+\left(5n-3\right)$$

20n^{2}-12n ichida 4n ni ajrating.

$$4n\left(5n-3\right)+5n-3$$

Distributiv funktsiyasidan foydalangan holda 5n-3 umumiy terminini chiqaring.

$$\left(5n-3\right)\left(4n+1\right)$$

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.

$$20n^{2}-7n-3=0$$

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.

$$n=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 20\left(-3\right)}}{2\times 20}$$

-7 kvadratini chiqarish.

$$n=\frac{-\left(-7\right)±\sqrt{49-4\times 20\left(-3\right)}}{2\times 20}$$

-4 ni 20 marotabaga ko'paytirish.

$$n=\frac{-\left(-7\right)±\sqrt{49-80\left(-3\right)}}{2\times 20}$$

-80 ni -3 marotabaga ko'paytirish.

$$n=\frac{-\left(-7\right)±\sqrt{49+240}}{2\times 20}$$

49 ni 240 ga qo'shish.

$$n=\frac{-\left(-7\right)±\sqrt{289}}{2\times 20}$$

289 ning kvadrat ildizini chiqarish.

$$n=\frac{-\left(-7\right)±17}{2\times 20}$$

-7 ning teskarisi 7 ga teng.

$$n=\frac{7±17}{2\times 20}$$

2 ni 20 marotabaga ko'paytirish.

$$n=\frac{7±17}{40}$$

n=\frac{7±17}{40} tenglamasini yeching, bunda ± musbat. 7 ni 17 ga qo'shish.

$$n=\frac{24}{40}$$

\frac{24}{40} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=\frac{3}{5}$$

n=\frac{7±17}{40} tenglamasini yeching, bunda ± manfiy. 7 dan 17 ni ayirish.

$$n=-\frac{10}{40}$$

\frac{-10}{40} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=-\frac{1}{4}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{3}{5} ga va x_{2} uchun -\frac{1}{4} ga bo‘ling.

$$20n^{2}-7n-3=20\left(n-\frac{3}{5}\right)\left(n-\left(-\frac{1}{4}\right)\right)$$

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

$$20n^{2}-7n-3=20\left(n-\frac{3}{5}\right)\left(n+\frac{1}{4}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{5} ni n dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$20n^{2}-7n-3=20\times \frac{5n-3}{5}\left(n+\frac{1}{4}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{4} ni n ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$20n^{2}-7n-3=20\times \frac{5n-3}{5}\times \frac{4n+1}{4}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{5n-3}{5} ni \frac{4n+1}{4} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$20n^{2}-7n-3=20\times \frac{\left(5n-3\right)\left(4n+1\right)}{5\times 4}$$

5 ni 4 marotabaga ko'paytirish.

$$20n^{2}-7n-3=20\times \frac{\left(5n-3\right)\left(4n+1\right)}{20}$$

20 va 20 ichida eng katta umumiy 20 faktorini bekor qiling.

$$20n^{2}-7n-3=\left(5n-3\right)\left(4n+1\right)$$

Baholash

$$\left(5n-3\right)\left(4n+1\right)$$