4(a+1)^2-4a^2+20geq0 eching

$$4 ( a + 1 ) ^ { 2 } - 4 a ^ { 2 } + 20 \geq 0$$

a uchun yechish

$$a\geq -3$$

Yechim qadamlari

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

$$4\left(a^{2}+2a+1\right)-4a^{2}+20\geq 0$$

4 ga a^{2}+2a+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4a^{2}+8a+4-4a^{2}+20\geq 0$$

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

$$8a+4+20\geq 0$$

24 olish uchun 4 va 20'ni qo'shing.

$$8a+24\geq 0$$

Ikkala tarafdan 24 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$8a\geq -24$$

Ikki tarafini 8 ga bo‘ling. 8 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$a\geq \frac{-24}{8}$$

-3 ni olish uchun -24 ni 8 ga bo‘ling.

$$a\geq -3$$