2p^2+4p-5=0 eching

$$2 p ^ { 2 } + 4 p - 5 = 0$$

p uchun yechish

$$p=\frac{\sqrt{14}}{2}-1\approx 0,870828693$$

$$p=-\frac{\sqrt{14}}{2}-1\approx -2,870828693$$

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.

$$2p^{2}+4p-5=0$$

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

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

4 kvadratini chiqarish.

$$p=\frac{-4±\sqrt{16-4\times 2\left(-5\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$p=\frac{-4±\sqrt{16-8\left(-5\right)}}{2\times 2}$$

-8 ni -5 marotabaga ko'paytirish.

$$p=\frac{-4±\sqrt{16+40}}{2\times 2}$$

16 ni 40 ga qo'shish.

$$p=\frac{-4±\sqrt{56}}{2\times 2}$$

56 ning kvadrat ildizini chiqarish.

$$p=\frac{-4±2\sqrt{14}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$p=\frac{-4±2\sqrt{14}}{4}$$

p=\frac{-4±2\sqrt{14}}{4} tenglamasini yeching, bunda ± musbat. -4 ni 2\sqrt{14} ga qo'shish.

$$p=\frac{2\sqrt{14}-4}{4}$$

-4+2\sqrt{14} ni 4 ga bo'lish.

$$p=\frac{\sqrt{14}}{2}-1$$

p=\frac{-4±2\sqrt{14}}{4} tenglamasini yeching, bunda ± manfiy. -4 dan 2\sqrt{14} ni ayirish.

$$p=\frac{-2\sqrt{14}-4}{4}$$

-4-2\sqrt{14} ni 4 ga bo'lish.

$$p=-\frac{\sqrt{14}}{2}-1$$

Tenglama yechildi.

$$p=\frac{\sqrt{14}}{2}-1 p=-\frac{\sqrt{14}}{2}-1$$

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.

$$2p^{2}+4p-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -5 ayirilsa 0 qoladi.

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

0 dan -5 ni ayirish.

$$2p^{2}+4p=5$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2p^{2}+4p}{2}=\frac{5}{2}$$

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

$$p^{2}+\frac{4}{2}p=\frac{5}{2}$$

4 ni 2 ga bo'lish.

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

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

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

1 kvadratini chiqarish.

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

\frac{5}{2} ni 1 ga qo'shish.

$$p^{2}+2p+1=\frac{7}{2}$$

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

$$\left(p+1\right)^{2}=\frac{7}{2}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(p+1\right)^{2}}=\sqrt{\frac{7}{2}}$$

Qisqartirish.

$$p+1=\frac{\sqrt{14}}{2} p+1=-\frac{\sqrt{14}}{2}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$p=\frac{\sqrt{14}}{2}-1 p=-\frac{\sqrt{14}}{2}-1$$