p^2-4+p=0 eching

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

p uchun yechish

$$p = \frac{\sqrt{17} - 1}{2} \approx 1,561552813$$

$$p=\frac{-\sqrt{17}-1}{2}\approx -2,561552813$$

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.

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

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

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

1 kvadratini chiqarish.

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

-4 ni -4 marotabaga ko'paytirish.

$$p=\frac{-1±\sqrt{1+16}}{2}$$

1 ni 16 ga qo'shish.

$$p=\frac{-1±\sqrt{17}}{2}$$

p=\frac{-1±\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{17} ga qo'shish.

$$p=\frac{\sqrt{17}-1}{2}$$

p=\frac{-1±\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{17} ni ayirish.

$$p=\frac{-\sqrt{17}-1}{2}$$

Tenglama yechildi.

$$p=\frac{\sqrt{17}-1}{2} p=\frac{-\sqrt{17}-1}{2}$$

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.

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

4 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -4 ayirilsa 0 qoladi.

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

0 dan -4 ni ayirish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$p^{2}+p+\frac{1}{4}=4+\frac{1}{4}$$

4 ni \frac{1}{4} ga qo'shish.

$$p^{2}+p+\frac{1}{4}=\frac{17}{4}$$

p^{2}+p+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(p+\frac{1}{2}\right)^{2}=\frac{17}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(p+\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Qisqartirish.

$$p+\frac{1}{2}=\frac{\sqrt{17}}{2} p+\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$p=\frac{\sqrt{17}-1}{2} p=\frac{-\sqrt{17}-1}{2}$$