(a+10)^2=(a-d+10)(a+d+11) eching

$$( a + 10 ) ^ { 2 } = ( a - d + 10 ) ( a + d + 11 )$$

a uchun yechish

$$a=d^{2}+d-10$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(a+10\right)^{2} kengaytirilishi uchun ishlating.

$$a^{2}+20a+100=\left(a-d+10\right)\left(a+d+11\right)$$

a-d+10 ga a+d+11 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$a^{2}+20a+100=a^{2}+21a-d^{2}-d+110$$

Ikkala tarafdan a^{2} ni ayirish.

$$a^{2}+20a+100-a^{2}=21a-d^{2}-d+110$$

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

$$20a+100=21a-d^{2}-d+110$$

Ikkala tarafdan 21a ni ayirish.

$$20a+100-21a=-d^{2}-d+110$$

-a ni olish uchun 20a va -21a ni birlashtirish.

$$-a+100=-d^{2}-d+110$$

Ikkala tarafdan 100 ni ayirish.

$$-a=-d^{2}-d+110-100$$

10 olish uchun 110 dan 100 ni ayirish.

$$-a=-d^{2}-d+10$$

Tenglama standart shaklda.

$$-a=10-d-d^{2}$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-a}{-1}=\frac{10-d-d^{2}}{-1}$$

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

$$a=\frac{10-d-d^{2}}{-1}$$

-d^{2}-d+10 ni -1 ga bo'lish.

$$a=d^{2}+d-10$$

d uchun yechish (complex solution)

$$d=\frac{\sqrt{4a+41}-1}{2}$$

$$d=\frac{-\sqrt{4a+41}-1}{2}$$

d uchun yechish

$$d=\frac{\sqrt{4a+41}-1}{2}$$

$$d=\frac{-\sqrt{4a+41}-1}{2}\text{, }a\geq -\frac{41}{4}$$