(x+1)^2=-25(y-1) eching

$$( x + 1 ) ^ { 2 } = - 25 ( y - 1 )$$

y uchun yechish

$$y=-\frac{\left(x-4\right)\left(x+6\right)}{25}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$x^{2}+2x+1=-25\left(y-1\right)$$

-25 ga y-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{2}+2x+1=-25y+25$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-25y+25=x^{2}+2x+1$$

Ikkala tarafdan 25 ni ayirish.

$$-25y=x^{2}+2x+1-25$$

-24 olish uchun 1 dan 25 ni ayirish.

$$-25y=x^{2}+2x-24$$

Ikki tarafini -25 ga bo‘ling.

$$\frac{-25y}{-25}=\frac{\left(x-4\right)\left(x+6\right)}{-25}$$

-25 ga bo'lish -25 ga ko'paytirishni bekor qiladi.

$$y=\frac{\left(x-4\right)\left(x+6\right)}{-25}$$

\left(-4+x\right)\left(6+x\right) ni -25 ga bo'lish.

$$y=-\frac{\left(x-4\right)\left(x+6\right)}{25}$$

x uchun yechish (complex solution)

$$x=-5\sqrt{1-y}-1$$

$$x=5\sqrt{1-y}-1$$

x uchun yechish

$$x=-5\sqrt{1-y}-1$$

$$x=5\sqrt{1-y}-1\text{, }y\leq 1$$