4p^2-25p+21=-4 eching

$$4 p ^ { 2 } - 25 p + 21 = - 4$$

p uchun yechish

$$p = \frac{5}{4} = 1\frac{1}{4} = 1,25$$

$$p=5$$

Guruhlash orqali faktorlash qadamlari

4 ni ikki tarafga qo’shing.

$$4p^{2}-25p+21+4=0$$

25 olish uchun 21 va 4'ni qo'shing.

$$4p^{2}-25p+25=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 4p^{2}+ap+bp+25 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-25 ab=4\times 25=100$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 100-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

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

Har bir juftlik yigʻindisini hisoblang.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

Yechim – -25 yigʻindisini beruvchi juftlik.

$$a=-20 b=-5$$

4p^{2}-25p+25 ni \left(4p^{2}-20p\right)+\left(-5p+25\right) sifatida qaytadan yozish.

$$\left(4p^{2}-20p\right)+\left(-5p+25\right)$$

Birinchi guruhda 4p ni va ikkinchi guruhda -5 ni faktordan chiqaring.

$$4p\left(p-5\right)-5\left(p-5\right)$$

Distributiv funktsiyasidan foydalangan holda p-5 umumiy terminini chiqaring.

$$\left(p-5\right)\left(4p-5\right)$$

Tenglamani yechish uchun p-5=0 va 4p-5=0 ni yeching.

$$p=5 p=\frac{5}{4}$$

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.

$$4p^{2}-25p+21=-4$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$4p^{2}-25p+21-\left(-4\right)=-4-\left(-4\right)$$

O‘zidan -4 ayirilsa 0 qoladi.

$$4p^{2}-25p+21-\left(-4\right)=0$$

21 dan -4 ni ayirish.

$$4p^{2}-25p+25=0$$

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

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

-25 kvadratini chiqarish.

$$p=\frac{-\left(-25\right)±\sqrt{625-4\times 4\times 25}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$p=\frac{-\left(-25\right)±\sqrt{625-16\times 25}}{2\times 4}$$

-16 ni 25 marotabaga ko'paytirish.

$$p=\frac{-\left(-25\right)±\sqrt{625-400}}{2\times 4}$$

625 ni -400 ga qo'shish.

$$p=\frac{-\left(-25\right)±\sqrt{225}}{2\times 4}$$

225 ning kvadrat ildizini chiqarish.

$$p=\frac{-\left(-25\right)±15}{2\times 4}$$

-25 ning teskarisi 25 ga teng.

$$p=\frac{25±15}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$p=\frac{25±15}{8}$$

p=\frac{25±15}{8} tenglamasini yeching, bunda ± musbat. 25 ni 15 ga qo'shish.

$$p=\frac{40}{8}$$

40 ni 8 ga bo'lish.

$$p=5$$

p=\frac{25±15}{8} tenglamasini yeching, bunda ± manfiy. 25 dan 15 ni ayirish.

$$p=\frac{10}{8}$$

\frac{10}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$p=\frac{5}{4}$$

Tenglama yechildi.

$$p=5 p=\frac{5}{4}$$

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.

$$4p^{2}-25p+21=-4$$

Tenglamaning ikkala tarafidan 21 ni ayirish.

$$4p^{2}-25p+21-21=-4-21$$

O‘zidan 21 ayirilsa 0 qoladi.

$$4p^{2}-25p=-4-21$$

-4 dan 21 ni ayirish.

$$4p^{2}-25p=-25$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4p^{2}-25p}{4}=-\frac{25}{4}$$

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

$$p^{2}-\frac{25}{4}p=-\frac{25}{4}$$

-\frac{25}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{25}{8} olish uchun. Keyin, -\frac{25}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$p^{2}-\frac{25}{4}p+\left(-\frac{25}{8}\right)^{2}=-\frac{25}{4}+\left(-\frac{25}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{25}{8} kvadratini chiqarish.

$$p^{2}-\frac{25}{4}p+\frac{625}{64}=-\frac{25}{4}+\frac{625}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{25}{4} ni \frac{625}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$p^{2}-\frac{25}{4}p+\frac{625}{64}=\frac{225}{64}$$

p^{2}-\frac{25}{4}p+\frac{625}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(p-\frac{25}{8}\right)^{2}=\frac{225}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(p-\frac{25}{8}\right)^{2}}=\sqrt{\frac{225}{64}}$$

Qisqartirish.

$$p-\frac{25}{8}=\frac{15}{8} p-\frac{25}{8}=-\frac{15}{8}$$

\frac{25}{8} ni tenglamaning ikkala tarafiga qo'shish.

$$p=5 p=\frac{5}{4}$$