25y^2-33y+8 eching

$$25 y ^ { 2 } - 33 y + 8$$

Omil

$$\left(y-1\right)\left(25y-8\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 25y^{2}+ay+by+8 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-33 ab=25\times 8=200$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 200-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-200 -2,-100 -4,-50 -5,-40 -8,-25 -10,-20$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-200=-201 -2-100=-102 -4-50=-54 -5-40=-45 -8-25=-33 -10-20=-30$$

Yechim – -33 yigʻindisini beruvchi juftlik.

$$a=-25 b=-8$$

25y^{2}-33y+8 ni \left(25y^{2}-25y\right)+\left(-8y+8\right) sifatida qaytadan yozish.

$$\left(25y^{2}-25y\right)+\left(-8y+8\right)$$

Birinchi guruhda 25y ni va ikkinchi guruhda -8 ni faktordan chiqaring.

$$25y\left(y-1\right)-8\left(y-1\right)$$

Distributiv funktsiyasidan foydalangan holda y-1 umumiy terminini chiqaring.

$$\left(y-1\right)\left(25y-8\right)$$

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.

$$25y^{2}-33y+8=0$$

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.

$$y=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 25\times 8}}{2\times 25}$$

-33 kvadratini chiqarish.

$$y=\frac{-\left(-33\right)±\sqrt{1089-4\times 25\times 8}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$y=\frac{-\left(-33\right)±\sqrt{1089-100\times 8}}{2\times 25}$$

-100 ni 8 marotabaga ko'paytirish.

$$y=\frac{-\left(-33\right)±\sqrt{1089-800}}{2\times 25}$$

1089 ni -800 ga qo'shish.

$$y=\frac{-\left(-33\right)±\sqrt{289}}{2\times 25}$$

289 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-33\right)±17}{2\times 25}$$

-33 ning teskarisi 33 ga teng.

$$y=\frac{33±17}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$y=\frac{33±17}{50}$$

y=\frac{33±17}{50} tenglamasini yeching, bunda ± musbat. 33 ni 17 ga qo'shish.

$$y=\frac{50}{50}$$

50 ni 50 ga bo'lish.

$$y=1$$

y=\frac{33±17}{50} tenglamasini yeching, bunda ± manfiy. 33 dan 17 ni ayirish.

$$y=\frac{16}{50}$$

\frac{16}{50} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=\frac{8}{25}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 1 ga va x_{2} uchun \frac{8}{25} ga bo‘ling.

$$25y^{2}-33y+8=25\left(y-1\right)\left(y-\frac{8}{25}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{8}{25} ni y dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$25y^{2}-33y+8=25\left(y-1\right)\times \frac{25y-8}{25}$$

25 va 25 ichida eng katta umumiy 25 faktorini bekor qiling.

$$25y^{2}-33y+8=\left(y-1\right)\left(25y-8\right)$$

Baholash

$$\left(y-1\right)\left(25y-8\right)$$