2y^2-9y-18 eching

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

Omil

$$\left(y-6\right)\left(2y+3\right)$$

Yechim qadamlari

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

$$a+b=-9 ab=2\left(-18\right)=-36$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -36-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

Yechim – -9 yigʻindisini beruvchi juftlik.

$$a=-12 b=3$$

2y^{2}-9y-18 ni \left(2y^{2}-12y\right)+\left(3y-18\right) sifatida qaytadan yozish.

$$\left(2y^{2}-12y\right)+\left(3y-18\right)$$

Birinchi guruhda 2y ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$2y\left(y-6\right)+3\left(y-6\right)$$

Distributiv funktsiyasidan foydalangan holda y-6 umumiy terminini chiqaring.

$$\left(y-6\right)\left(2y+3\right)$$

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.

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

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.

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

-9 kvadratini chiqarish.

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

-4 ni 2 marotabaga ko'paytirish.

$$y=\frac{-\left(-9\right)±\sqrt{81-8\left(-18\right)}}{2\times 2}$$

-8 ni -18 marotabaga ko'paytirish.

$$y=\frac{-\left(-9\right)±\sqrt{81+144}}{2\times 2}$$

81 ni 144 ga qo'shish.

$$y=\frac{-\left(-9\right)±\sqrt{225}}{2\times 2}$$

225 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-9\right)±15}{2\times 2}$$

-9 ning teskarisi 9 ga teng.

$$y=\frac{9±15}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$y=\frac{9±15}{4}$$

y=\frac{9±15}{4} tenglamasini yeching, bunda ± musbat. 9 ni 15 ga qo'shish.

$$y=\frac{24}{4}$$

24 ni 4 ga bo'lish.

$$y=6$$

y=\frac{9±15}{4} tenglamasini yeching, bunda ± manfiy. 9 dan 15 ni ayirish.

$$y=-\frac{6}{4}$$

\frac{-6}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=-\frac{3}{2}$$

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

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

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

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{2} ni y ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

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

2 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

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

Baholash

$$\left(y-6\right)\left(2y+3\right)$$