5y^2-9y-18 eching

$$5 y ^ { 2 } - 9 y - 18$$

Omil

$$\left(y-3\right)\left(5y+6\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 5y^{2}+ay+by-18 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-9 ab=5\left(-18\right)=-90$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -90-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

Yechim – -9 yigʻindisini beruvchi juftlik.

$$a=-15 b=6$$

5y^{2}-9y-18 ni \left(5y^{2}-15y\right)+\left(6y-18\right) sifatida qaytadan yozish.

$$\left(5y^{2}-15y\right)+\left(6y-18\right)$$

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

$$5y\left(y-3\right)+6\left(y-3\right)$$

Distributiv funktsiyasidan foydalangan holda y-3 umumiy terminini chiqaring.

$$\left(y-3\right)\left(5y+6\right)$$

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.

$$5y^{2}-9y-18=0$$

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.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-18\right)}}{2\times 5}$$

-9 kvadratini chiqarish.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\times 5\left(-18\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$y=\frac{-\left(-9\right)±\sqrt{81-20\left(-18\right)}}{2\times 5}$$

-20 ni -18 marotabaga ko'paytirish.

$$y=\frac{-\left(-9\right)±\sqrt{81+360}}{2\times 5}$$

81 ni 360 ga qo'shish.

$$y=\frac{-\left(-9\right)±\sqrt{441}}{2\times 5}$$

441 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-9\right)±21}{2\times 5}$$

-9 ning teskarisi 9 ga teng.

$$y=\frac{9±21}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$y=\frac{9±21}{10}$$

y=\frac{9±21}{10} tenglamasini yeching, bunda ± musbat. 9 ni 21 ga qo'shish.

$$y=\frac{30}{10}$$

30 ni 10 ga bo'lish.

$$y=3$$

y=\frac{9±21}{10} tenglamasini yeching, bunda ± manfiy. 9 dan 21 ni ayirish.

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

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

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 3 ga va x_{2} uchun -\frac{6}{5} ga bo‘ling.

$$5y^{2}-9y-18=5\left(y-3\right)\left(y-\left(-\frac{6}{5}\right)\right)$$

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

$$5y^{2}-9y-18=5\left(y-3\right)\left(y+\frac{6}{5}\right)$$

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

$$5y^{2}-9y-18=5\left(y-3\right)\times \frac{5y+6}{5}$$

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

$$5y^{2}-9y-18=\left(y-3\right)\left(5y+6\right)$$

Baholash

$$\left(y-3\right)\left(5y+6\right)$$