6y^2+5y-25 eching

$$6 y ^ { 2 } + 5 y - 25$$

Omil

$$\left(3y-5\right)\left(2y+5\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6y^{2}+ay+by-25 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=5 ab=6\left(-25\right)=-150$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -150-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,150 -2,75 -3,50 -5,30 -6,25 -10,15$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+150=149 -2+75=73 -3+50=47 -5+30=25 -6+25=19 -10+15=5$$

Yechim – 5 yigʻindisini beruvchi juftlik.

$$a=-10 b=15$$

6y^{2}+5y-25 ni \left(6y^{2}-10y\right)+\left(15y-25\right) sifatida qaytadan yozish.

$$\left(6y^{2}-10y\right)+\left(15y-25\right)$$

Birinchi guruhda 2y ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$2y\left(3y-5\right)+5\left(3y-5\right)$$

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

$$\left(3y-5\right)\left(2y+5\right)$$

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.

$$6y^{2}+5y-25=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}-4\times 6\left(-25\right)}}{2\times 6}$$

5 kvadratini chiqarish.

$$y=\frac{-5±\sqrt{25-4\times 6\left(-25\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$y=\frac{-5±\sqrt{25-24\left(-25\right)}}{2\times 6}$$

-24 ni -25 marotabaga ko'paytirish.

$$y=\frac{-5±\sqrt{25+600}}{2\times 6}$$

25 ni 600 ga qo'shish.

$$y=\frac{-5±\sqrt{625}}{2\times 6}$$

625 ning kvadrat ildizini chiqarish.

$$y=\frac{-5±25}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$y=\frac{-5±25}{12}$$

y=\frac{-5±25}{12} tenglamasini yeching, bunda ± musbat. -5 ni 25 ga qo'shish.

$$y=\frac{20}{12}$$

\frac{20}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

y=\frac{-5±25}{12} tenglamasini yeching, bunda ± manfiy. -5 dan 25 ni ayirish.

$$y=-\frac{30}{12}$$

\frac{-30}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=-\frac{5}{2}$$

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

$$6y^{2}+5y-25=6\left(y-\frac{5}{3}\right)\left(y-\left(-\frac{5}{2}\right)\right)$$

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

$$6y^{2}+5y-25=6\left(y-\frac{5}{3}\right)\left(y+\frac{5}{2}\right)$$

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

$$6y^{2}+5y-25=6\times \frac{3y-5}{3}\left(y+\frac{5}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni y ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6y^{2}+5y-25=6\times \frac{3y-5}{3}\times \frac{2y+5}{2}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{3y-5}{3} ni \frac{2y+5}{2} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$6y^{2}+5y-25=6\times \frac{\left(3y-5\right)\left(2y+5\right)}{3\times 2}$$

3 ni 2 marotabaga ko'paytirish.

$$6y^{2}+5y-25=6\times \frac{\left(3y-5\right)\left(2y+5\right)}{6}$$

6 va 6 ichida eng katta umumiy 6 faktorini bekor qiling.

$$6y^{2}+5y-25=\left(3y-5\right)\left(2y+5\right)$$

Baholash

$$\left(3y-5\right)\left(2y+5\right)$$