p-1=sqrt{50-2p} eching

$$p - 1 = \sqrt { 50 - 2 p }$$

p uchun yechish

$$p=7$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(p-1\right)^{2}=\left(\sqrt{50-2p}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(p-1\right)^{2} kengaytirilishi uchun ishlating.

$$p^{2}-2p+1=\left(\sqrt{50-2p}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{50-2p} ga hisoblang va 50-2p ni qiymatni oling.

$$p^{2}-2p+1=50-2p$$

Ikkala tarafdan 50 ni ayirish.

$$p^{2}-2p+1-50=-2p$$

-49 olish uchun 1 dan 50 ni ayirish.

$$p^{2}-2p-49=-2p$$

2p ni ikki tarafga qo’shing.

$$p^{2}-2p-49+2p=0$$

0 ni olish uchun -2p va 2p ni birlashtirish.

$$p^{2}-49=0$$

Hisoblang: p^{2}-49. p^{2}-49 ni p^{2}-7^{2} sifatida qaytadan yozish. Kvadratlarning farqini ushbu formula bilan hisoblash mumkin: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(p-7\right)\left(p+7\right)=0$$

Tenglamani yechish uchun p-7=0 va p+7=0 ni yeching.

$$p=7 p=-7$$

p-1=\sqrt{50-2p} tenglamasida p uchun 7 ni almashtiring.

$$7-1=\sqrt{50-2\times 7}$$

Qisqartirish. p=7 tenglamani qoniqtiradi.

$$6=6$$

p-1=\sqrt{50-2p} tenglamasida p uchun -7 ni almashtiring.

$$-7-1=\sqrt{50-2\left(-7\right)}$$

Qisqartirish. p=-7 qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$-8=8$$

p-1=\sqrt{50-2p} tenglamasi noyob yechimga ega.

$$p=7$$