2p^2+p-5=0 eching

$$2 p ^ { 2 } + p - 5 = 0$$

p uchun yechish

$$p = \frac{\sqrt{41} - 1}{4} \approx 1,350781059$$

$$p=\frac{-\sqrt{41}-1}{4}\approx -1,850781059$$

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.

$$2p^{2}+p-5=0$$

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 -5 ni c bilan almashtiring.

$$p=\frac{-1±\sqrt{1^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

1 kvadratini chiqarish.

$$p=\frac{-1±\sqrt{1-4\times 2\left(-5\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$p=\frac{-1±\sqrt{1-8\left(-5\right)}}{2\times 2}$$

-8 ni -5 marotabaga ko'paytirish.

$$p=\frac{-1±\sqrt{1+40}}{2\times 2}$$

1 ni 40 ga qo'shish.

$$p=\frac{-1±\sqrt{41}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$p=\frac{-1±\sqrt{41}}{4}$$

p=\frac{-1±\sqrt{41}}{4} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{41} ga qo'shish.

$$p=\frac{\sqrt{41}-1}{4}$$

p=\frac{-1±\sqrt{41}}{4} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{41} ni ayirish.

$$p=\frac{-\sqrt{41}-1}{4}$$

Tenglama yechildi.

$$p=\frac{\sqrt{41}-1}{4} p=\frac{-\sqrt{41}-1}{4}$$

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.

$$2p^{2}+p-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -5 ayirilsa 0 qoladi.

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

0 dan -5 ni ayirish.

$$2p^{2}+p=5$$

Ikki tarafini 2 ga bo‘ling.

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

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

$$p^{2}+\frac{1}{2}p=\frac{5}{2}$$

\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.

$$p^{2}+\frac{1}{2}p+\left(\frac{1}{4}\right)^{2}=\frac{5}{2}+\left(\frac{1}{4}\right)^{2}$$

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

$$p^{2}+\frac{1}{2}p+\frac{1}{16}=\frac{5}{2}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$p^{2}+\frac{1}{2}p+\frac{1}{16}=\frac{41}{16}$$

p^{2}+\frac{1}{2}p+\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.

$$\left(p+\frac{1}{4}\right)^{2}=\frac{41}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(p+\frac{1}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Qisqartirish.

$$p+\frac{1}{4}=\frac{\sqrt{41}}{4} p+\frac{1}{4}=-\frac{\sqrt{41}}{4}$$

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

$$p=\frac{\sqrt{41}-1}{4} p=\frac{-\sqrt{41}-1}{4}$$