(A):P(A)=1:2f eching

$$( A ) : P ( A ) = 1 : 2 f$$

P uchun yechish

$$\left\{\begin{matrix}P=\frac{2A^{2}}{f}\text{, }&A\neq 0\text{ and }f\neq 0\\P\neq 0\text{, }&f=0\text{ and }A=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

P qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2P ga, P,2 ning eng kichik karralisiga ko‘paytiring.

$$2AA=\frac{1}{2}f\times 2P$$

A^{2} hosil qilish uchun A va A ni ko'paytirish.

$$2A^{2}=\frac{1}{2}f\times 2P$$

1 hosil qilish uchun \frac{1}{2} va 2 ni ko'paytirish.

$$2A^{2}=fP$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$fP=2A^{2}$$

Ikki tarafini f ga bo‘ling.

$$\frac{fP}{f}=\frac{2A^{2}}{f}$$

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

$$P=\frac{2A^{2}}{f}$$

P qiymati 0 teng bo‘lmaydi.

$$P=\frac{2A^{2}}{f}\text{, }P\neq 0$$

A uchun yechish (complex solution)

$$A=-\frac{\sqrt{P}\sqrt{2f}}{2}$$

$$A=\frac{\sqrt{P}\sqrt{2f}}{2}\text{, }P\neq 0$$

A uchun yechish

$$A=\frac{\sqrt{2Pf}}{2}$$

$$A=-\frac{\sqrt{2Pf}}{2}\text{, }\left(f\geq 0\text{ and }P>0\right)\text{ or }\left(f\leq 0\text{ and }P<0\right)$$