6p^2-5=13p eching

$$6 p ^ { 2 } - 5 = 13 p$$

p uchun yechish

$$p=-\frac{1}{3}\approx -0,333333333$$

$$p = \frac{5}{2} = 2\frac{1}{2} = 2,5$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan 13p ni ayirish.

$$6p^{2}-5-13p=0$$

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$6p^{2}-13p-5=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 6p^{2}+ap+bp-5 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-13 ab=6\left(-5\right)=-30$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -30-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-30 2,-15 3,-10 5,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

Yechim – -13 yigʻindisini beruvchi juftlik.

$$a=-15 b=2$$

6p^{2}-13p-5 ni \left(6p^{2}-15p\right)+\left(2p-5\right) sifatida qaytadan yozish.

$$\left(6p^{2}-15p\right)+\left(2p-5\right)$$

6p^{2}-15p ichida 3p ni ajrating.

$$3p\left(2p-5\right)+2p-5$$

Distributiv funktsiyasidan foydalangan holda 2p-5 umumiy terminini chiqaring.

$$\left(2p-5\right)\left(3p+1\right)$$

Tenglamani yechish uchun 2p-5=0 va 3p+1=0 ni yeching.

$$p=\frac{5}{2} p=-\frac{1}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 13p ni ayirish.

$$6p^{2}-5-13p=0$$

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.

$$6p^{2}-13p-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 6 ni a, -13 ni b va -5 ni c bilan almashtiring.

$$p=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

-13 kvadratini chiqarish.

$$p=\frac{-\left(-13\right)±\sqrt{169-4\times 6\left(-5\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$p=\frac{-\left(-13\right)±\sqrt{169-24\left(-5\right)}}{2\times 6}$$

-24 ni -5 marotabaga ko'paytirish.

$$p=\frac{-\left(-13\right)±\sqrt{169+120}}{2\times 6}$$

169 ni 120 ga qo'shish.

$$p=\frac{-\left(-13\right)±\sqrt{289}}{2\times 6}$$

289 ning kvadrat ildizini chiqarish.

$$p=\frac{-\left(-13\right)±17}{2\times 6}$$

-13 ning teskarisi 13 ga teng.

$$p=\frac{13±17}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$p=\frac{13±17}{12}$$

p=\frac{13±17}{12} tenglamasini yeching, bunda ± musbat. 13 ni 17 ga qo'shish.

$$p=\frac{30}{12}$$

\frac{30}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$p=\frac{5}{2}$$

p=\frac{13±17}{12} tenglamasini yeching, bunda ± manfiy. 13 dan 17 ni ayirish.

$$p=-\frac{4}{12}$$

\frac{-4}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$p=-\frac{1}{3}$$

Tenglama yechildi.

$$p=\frac{5}{2} p=-\frac{1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 13p ni ayirish.

$$6p^{2}-5-13p=0$$

5 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$6p^{2}-13p=5$$

Ikki tarafini 6 ga bo‘ling.

$$\frac{6p^{2}-13p}{6}=\frac{5}{6}$$

6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.

$$p^{2}-\frac{13}{6}p=\frac{5}{6}$$

-\frac{13}{6} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{13}{12} olish uchun. Keyin, -\frac{13}{12} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$p^{2}-\frac{13}{6}p+\left(-\frac{13}{12}\right)^{2}=\frac{5}{6}+\left(-\frac{13}{12}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{13}{12} kvadratini chiqarish.

$$p^{2}-\frac{13}{6}p+\frac{169}{144}=\frac{5}{6}+\frac{169}{144}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{6} ni \frac{169}{144} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$p^{2}-\frac{13}{6}p+\frac{169}{144}=\frac{289}{144}$$

p^{2}-\frac{13}{6}p+\frac{169}{144} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(p-\frac{13}{12}\right)^{2}=\frac{289}{144}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(p-\frac{13}{12}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Qisqartirish.

$$p-\frac{13}{12}=\frac{17}{12} p-\frac{13}{12}=-\frac{17}{12}$$

\frac{13}{12} ni tenglamaning ikkala tarafiga qo'shish.

$$p=\frac{5}{2} p=-\frac{1}{3}$$