4y^2-24y+27 eching

$$4 y ^ { 2 } - 24 y + 27$$

Omil

$$\left(2y-9\right)\left(2y-3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 4y^{2}+ay+by+27 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-24 ab=4\times 27=108$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 108-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-108 -2,-54 -3,-36 -4,-27 -6,-18 -9,-12$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-108=-109 -2-54=-56 -3-36=-39 -4-27=-31 -6-18=-24 -9-12=-21$$

Yechim – -24 yigʻindisini beruvchi juftlik.

$$a=-18 b=-6$$

4y^{2}-24y+27 ni \left(4y^{2}-18y\right)+\left(-6y+27\right) sifatida qaytadan yozish.

$$\left(4y^{2}-18y\right)+\left(-6y+27\right)$$

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

$$2y\left(2y-9\right)-3\left(2y-9\right)$$

Distributiv funktsiyasidan foydalangan holda 2y-9 umumiy terminini chiqaring.

$$\left(2y-9\right)\left(2y-3\right)$$

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.

$$4y^{2}-24y+27=0$$

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.

$$y=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 4\times 27}}{2\times 4}$$

-24 kvadratini chiqarish.

$$y=\frac{-\left(-24\right)±\sqrt{576-4\times 4\times 27}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$y=\frac{-\left(-24\right)±\sqrt{576-16\times 27}}{2\times 4}$$

-16 ni 27 marotabaga ko'paytirish.

$$y=\frac{-\left(-24\right)±\sqrt{576-432}}{2\times 4}$$

576 ni -432 ga qo'shish.

$$y=\frac{-\left(-24\right)±\sqrt{144}}{2\times 4}$$

144 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-24\right)±12}{2\times 4}$$

-24 ning teskarisi 24 ga teng.

$$y=\frac{24±12}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$y=\frac{24±12}{8}$$

y=\frac{24±12}{8} tenglamasini yeching, bunda ± musbat. 24 ni 12 ga qo'shish.

$$y=\frac{36}{8}$$

\frac{36}{8} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=\frac{9}{2}$$

y=\frac{24±12}{8} tenglamasini yeching, bunda ± manfiy. 24 dan 12 ni ayirish.

$$y=\frac{12}{8}$$

\frac{12}{8} ulushini 4 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 \frac{9}{2} ga va x_{2} uchun \frac{3}{2} ga bo‘ling.

$$4y^{2}-24y+27=4\left(y-\frac{9}{2}\right)\left(y-\frac{3}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{9}{2} ni y dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$4y^{2}-24y+27=4\times \frac{2y-9}{2}\left(y-\frac{3}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{2} ni y dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$4y^{2}-24y+27=4\times \frac{2y-9}{2}\times \frac{2y-3}{2}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{2y-9}{2} ni \frac{2y-3}{2} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$4y^{2}-24y+27=4\times \frac{\left(2y-9\right)\left(2y-3\right)}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$4y^{2}-24y+27=4\times \frac{\left(2y-9\right)\left(2y-3\right)}{4}$$

4 va 4 ichida eng katta umumiy 4 faktorini bekor qiling.

$$4y^{2}-24y+27=\left(2y-9\right)\left(2y-3\right)$$

Baholash

$$\left(2y-9\right)\left(2y-3\right)$$