25y^2-54y-63=0 eching

$$25 y ^ { 2 } - 54 y - 63 = 0$$

y uchun yechish

$$y=-\frac{21}{25}=-0,84$$

$$y=3$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 25y^{2}+ay+by-63 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-54 ab=25\left(-63\right)=-1575$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -1575-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-1575 3,-525 5,-315 7,-225 9,-175 15,-105 21,-75 25,-63 35,-45$$

Har bir juftlik yigʻindisini hisoblang.

$$1-1575=-1574 3-525=-522 5-315=-310 7-225=-218 9-175=-166 15-105=-90 21-75=-54 25-63=-38 35-45=-10$$

Yechim – -54 yigʻindisini beruvchi juftlik.

$$a=-75 b=21$$

25y^{2}-54y-63 ni \left(25y^{2}-75y\right)+\left(21y-63\right) sifatida qaytadan yozish.

$$\left(25y^{2}-75y\right)+\left(21y-63\right)$$

Birinchi guruhda 25y ni va ikkinchi guruhda 21 ni faktordan chiqaring.

$$25y\left(y-3\right)+21\left(y-3\right)$$

Distributiv funktsiyasidan foydalangan holda y-3 umumiy terminini chiqaring.

$$\left(y-3\right)\left(25y+21\right)$$

Tenglamani yechish uchun y-3=0 va 25y+21=0 ni yeching.

$$y=3 y=-\frac{21}{25}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$25y^{2}-54y-63=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 25 ni a, -54 ni b va -63 ni c bilan almashtiring.

$$y=\frac{-\left(-54\right)±\sqrt{\left(-54\right)^{2}-4\times 25\left(-63\right)}}{2\times 25}$$

-54 kvadratini chiqarish.

$$y=\frac{-\left(-54\right)±\sqrt{2916-4\times 25\left(-63\right)}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$y=\frac{-\left(-54\right)±\sqrt{2916-100\left(-63\right)}}{2\times 25}$$

-100 ni -63 marotabaga ko'paytirish.

$$y=\frac{-\left(-54\right)±\sqrt{2916+6300}}{2\times 25}$$

2916 ni 6300 ga qo'shish.

$$y=\frac{-\left(-54\right)±\sqrt{9216}}{2\times 25}$$

9216 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-54\right)±96}{2\times 25}$$

-54 ning teskarisi 54 ga teng.

$$y=\frac{54±96}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$y=\frac{54±96}{50}$$

y=\frac{54±96}{50} tenglamasini yeching, bunda ± musbat. 54 ni 96 ga qo'shish.

$$y=\frac{150}{50}$$

150 ni 50 ga bo'lish.

$$y=3$$

y=\frac{54±96}{50} tenglamasini yeching, bunda ± manfiy. 54 dan 96 ni ayirish.

$$y=-\frac{42}{50}$$

\frac{-42}{50} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=-\frac{21}{25}$$

Tenglama yechildi.

$$y=3 y=-\frac{21}{25}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$25y^{2}-54y-63=0$$

63 ni tenglamaning ikkala tarafiga qo'shish.

$$25y^{2}-54y-63-\left(-63\right)=-\left(-63\right)$$

O‘zidan -63 ayirilsa 0 qoladi.

$$25y^{2}-54y=-\left(-63\right)$$

0 dan -63 ni ayirish.

$$25y^{2}-54y=63$$

Ikki tarafini 25 ga bo‘ling.

$$\frac{25y^{2}-54y}{25}=\frac{63}{25}$$

25 ga bo'lish 25 ga ko'paytirishni bekor qiladi.

$$y^{2}-\frac{54}{25}y=\frac{63}{25}$$

-\frac{54}{25} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{27}{25} olish uchun. Keyin, -\frac{27}{25} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$y^{2}-\frac{54}{25}y+\left(-\frac{27}{25}\right)^{2}=\frac{63}{25}+\left(-\frac{27}{25}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{27}{25} kvadratini chiqarish.

$$y^{2}-\frac{54}{25}y+\frac{729}{625}=\frac{63}{25}+\frac{729}{625}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{63}{25} ni \frac{729}{625} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$y^{2}-\frac{54}{25}y+\frac{729}{625}=\frac{2304}{625}$$

y^{2}-\frac{54}{25}y+\frac{729}{625} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(y-\frac{27}{25}\right)^{2}=\frac{2304}{625}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-\frac{27}{25}\right)^{2}}=\sqrt{\frac{2304}{625}}$$

Qisqartirish.

$$y-\frac{27}{25}=\frac{48}{25} y-\frac{27}{25}=-\frac{48}{25}$$

\frac{27}{25} ni tenglamaning ikkala tarafiga qo'shish.

$$y=3 y=-\frac{21}{25}$$