p^2+5/6=p eching

$$\frac { p ^ { 2 } + 5 } { 6 } = p$$

p uchun yechish

$$p=1$$

$$p=5$$

Kvadrat tenglamalaridan foydalanish qoidalari

\frac{1}{6}p^{2}+\frac{5}{6} natijani olish uchun p^{2}+5 ning har bir ifodasini 6 ga bo‘ling.

$$\frac{1}{6}p^{2}+\frac{5}{6}=p$$

Ikkala tarafdan p ni ayirish.

$$\frac{1}{6}p^{2}+\frac{5}{6}-p=0$$

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.

$$\frac{1}{6}p^{2}-p+\frac{5}{6}=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} \frac{1}{6} ni a, -1 ni b va \frac{5}{6} ni c bilan almashtiring.

$$p=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{6}\times \frac{5}{6}}}{2\times \frac{1}{6}}$$

-4 ni \frac{1}{6} marotabaga ko'paytirish.

$$p=\frac{-\left(-1\right)±\sqrt{1-\frac{2}{3}\times \frac{5}{6}}}{2\times \frac{1}{6}}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali -\frac{2}{3} ni \frac{5}{6} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$p=\frac{-\left(-1\right)±\sqrt{1-\frac{5}{9}}}{2\times \frac{1}{6}}$$

1 ni -\frac{5}{9} ga qo'shish.

$$p=\frac{-\left(-1\right)±\sqrt{\frac{4}{9}}}{2\times \frac{1}{6}}$$

\frac{4}{9} ning kvadrat ildizini chiqarish.

$$p=\frac{-\left(-1\right)±\frac{2}{3}}{2\times \frac{1}{6}}$$

-1 ning teskarisi 1 ga teng.

$$p=\frac{1±\frac{2}{3}}{2\times \frac{1}{6}}$$

2 ni \frac{1}{6} marotabaga ko'paytirish.

$$p=\frac{1±\frac{2}{3}}{\frac{1}{3}}$$

p=\frac{1±\frac{2}{3}}{\frac{1}{3}} tenglamasini yeching, bunda ± musbat. 1 ni \frac{2}{3} ga qo'shish.

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

\frac{5}{3} ni \frac{1}{3} ga bo'lish \frac{5}{3} ga k'paytirish \frac{1}{3} ga qaytarish.

$$p=5$$

p=\frac{1±\frac{2}{3}}{\frac{1}{3}} tenglamasini yeching, bunda ± manfiy. 1 dan \frac{2}{3} ni ayirish.

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

\frac{1}{3} ni \frac{1}{3} ga bo'lish \frac{1}{3} ga k'paytirish \frac{1}{3} ga qaytarish.

$$p=1$$

Tenglama yechildi.

$$p=5 p=1$$

Kvadratni yakunlash bo'yicha qadamlar

\frac{1}{6}p^{2}+\frac{5}{6} natijani olish uchun p^{2}+5 ning har bir ifodasini 6 ga bo‘ling.

$$\frac{1}{6}p^{2}+\frac{5}{6}=p$$

Ikkala tarafdan p ni ayirish.

$$\frac{1}{6}p^{2}+\frac{5}{6}-p=0$$

Ikkala tarafdan \frac{5}{6} ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$\frac{1}{6}p^{2}-p=-\frac{5}{6}$$

Ikkala tarafini 6 ga ko‘paytiring.

$$\frac{\frac{1}{6}p^{2}-p}{\frac{1}{6}}=-\frac{\frac{5}{6}}{\frac{1}{6}}$$

\frac{1}{6} ga bo'lish \frac{1}{6} ga ko'paytirishni bekor qiladi.

$$p^{2}+\left(-\frac{1}{\frac{1}{6}}\right)p=-\frac{\frac{5}{6}}{\frac{1}{6}}$$

-1 ni \frac{1}{6} ga bo'lish -1 ga k'paytirish \frac{1}{6} ga qaytarish.

$$p^{2}-6p=-\frac{\frac{5}{6}}{\frac{1}{6}}$$

-\frac{5}{6} ni \frac{1}{6} ga bo'lish -\frac{5}{6} ga k'paytirish \frac{1}{6} ga qaytarish.

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

-6 ni bo‘lish, x shartining koeffitsienti, 2 ga -3 olish uchun. Keyin, -3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$p^{2}-6p+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

-3 kvadratini chiqarish.

$$p^{2}-6p+9=-5+9$$

-5 ni 9 ga qo'shish.

$$p^{2}-6p+9=4$$

p^{2}-6p+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(p-3\right)^{2}=4$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(p-3\right)^{2}}=\sqrt{4}$$

Qisqartirish.

$$p-3=2 p-3=-2$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$p=5 p=1$$