4x^2+12x=4(x+p)^2-q eching

$$4 x ^ { 2 } + 12 x = 4 ( x + p ) ^ { 2 } - q$$

q uchun yechish

$$q=4\left(p\left(2x+p\right)-3x\right)$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$4x^{2}+12x=4\left(x^{2}+2xp+p^{2}\right)-q$$

4 ga x^{2}+2xp+p^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4x^{2}+12x=4x^{2}+8xp+4p^{2}-q$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$4x^{2}+8xp+4p^{2}-q=4x^{2}+12x$$

Ikkala tarafdan 4x^{2} ni ayirish.

$$8xp+4p^{2}-q=4x^{2}+12x-4x^{2}$$

0 ni olish uchun 4x^{2} va -4x^{2} ni birlashtirish.

$$8xp+4p^{2}-q=12x$$

Ikkala tarafdan 8xp ni ayirish.

$$4p^{2}-q=12x-8xp$$

Ikkala tarafdan 4p^{2} ni ayirish.

$$-q=12x-8xp-4p^{2}$$

Tenglama standart shaklda.

$$-q=-8px+12x-4p^{2}$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-q}{-1}=\frac{-8px+12x-4p^{2}}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$q=\frac{-8px+12x-4p^{2}}{-1}$$

12x-8xp-4p^{2} ni -1 ga bo'lish.

$$q=8px-12x+4p^{2}$$

p uchun yechish (complex solution)

$$p=-\frac{\sqrt{4x^{2}+12x+q}}{2}-x$$

$$p=\frac{\sqrt{4x^{2}+12x+q}}{2}-x$$

p uchun yechish

$$p=-\frac{\sqrt{4x^{2}+12x+q}}{2}-x$$

$$p=\frac{\sqrt{4x^{2}+12x+q}}{2}-x\text{, }q\geq -4x^{2}-12x$$