2w^2+w-1275=0 eching

$$2 w ^ { 2 } + w - 1275 = 0$$

w uchun yechish

$$w = -\frac{51}{2} = -25\frac{1}{2} = -25,5$$

$$w=25$$

Guruhlash orqali faktorlash qadamlari

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 2w^{2}+aw+bw-1275 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=1 ab=2\left(-1275\right)=-2550$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -2550-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,2550 -2,1275 -3,850 -5,510 -6,425 -10,255 -15,170 -17,150 -25,102 -30,85 -34,75 -50,51$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+2550=2549 -2+1275=1273 -3+850=847 -5+510=505 -6+425=419 -10+255=245 -15+170=155 -17+150=133 -25+102=77 -30+85=55 -34+75=41 -50+51=1$$

Yechim – 1 yigʻindisini beruvchi juftlik.

$$a=-50 b=51$$

2w^{2}+w-1275 ni \left(2w^{2}-50w\right)+\left(51w-1275\right) sifatida qaytadan yozish.

$$\left(2w^{2}-50w\right)+\left(51w-1275\right)$$

Birinchi guruhda 2w ni va ikkinchi guruhda 51 ni faktordan chiqaring.

$$2w\left(w-25\right)+51\left(w-25\right)$$

Distributiv funktsiyasidan foydalangan holda w-25 umumiy terminini chiqaring.

$$\left(w-25\right)\left(2w+51\right)$$

Tenglamani yechish uchun w-25=0 va 2w+51=0 ni yeching.

$$w=25 w=-\frac{51}{2}$$

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.

$$2w^{2}+w-1275=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 1 ni b va -1275 ni c bilan almashtiring.

$$w=\frac{-1±\sqrt{1^{2}-4\times 2\left(-1275\right)}}{2\times 2}$$

1 kvadratini chiqarish.

$$w=\frac{-1±\sqrt{1-4\times 2\left(-1275\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$w=\frac{-1±\sqrt{1-8\left(-1275\right)}}{2\times 2}$$

-8 ni -1275 marotabaga ko'paytirish.

$$w=\frac{-1±\sqrt{1+10200}}{2\times 2}$$

1 ni 10200 ga qo'shish.

$$w=\frac{-1±\sqrt{10201}}{2\times 2}$$

10201 ning kvadrat ildizini chiqarish.

$$w=\frac{-1±101}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$w=\frac{-1±101}{4}$$

w=\frac{-1±101}{4} tenglamasini yeching, bunda ± musbat. -1 ni 101 ga qo'shish.

$$w=\frac{100}{4}$$

100 ni 4 ga bo'lish.

$$w=25$$

w=\frac{-1±101}{4} tenglamasini yeching, bunda ± manfiy. -1 dan 101 ni ayirish.

$$w=-\frac{102}{4}$$

\frac{-102}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$w=-\frac{51}{2}$$

Tenglama yechildi.

$$w=25 w=-\frac{51}{2}$$

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.

$$2w^{2}+w-1275=0$$

1275 ni tenglamaning ikkala tarafiga qo'shish.

$$2w^{2}+w-1275-\left(-1275\right)=-\left(-1275\right)$$

O‘zidan -1275 ayirilsa 0 qoladi.

$$2w^{2}+w=-\left(-1275\right)$$

0 dan -1275 ni ayirish.

$$2w^{2}+w=1275$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2w^{2}+w}{2}=\frac{1275}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$w^{2}+\frac{1}{2}w=\frac{1275}{2}$$

\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{4} olish uchun. Keyin, \frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$w^{2}+\frac{1}{2}w+\left(\frac{1}{4}\right)^{2}=\frac{1275}{2}+\left(\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$w^{2}+\frac{1}{2}w+\frac{1}{16}=\frac{1275}{2}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1275}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$w^{2}+\frac{1}{2}w+\frac{1}{16}=\frac{10201}{16}$$

w^{2}+\frac{1}{2}w+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(w+\frac{1}{4}\right)^{2}=\frac{10201}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(w+\frac{1}{4}\right)^{2}}=\sqrt{\frac{10201}{16}}$$

Qisqartirish.

$$w+\frac{1}{4}=\frac{101}{4} w+\frac{1}{4}=-\frac{101}{4}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$w=25 w=-\frac{51}{2}$$