(2x+1)^2-5=0 eching

$$( 2 x + 1 ) ^ { 2 } - 5 = 0$$

x uchun yechish

$$x=\frac{\sqrt{5}-1}{2}\approx 0,618033989$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1,618033989$$

Kvadrat ildizini topish bo'yicha qadamlar

5 ni tenglamaning ikkala tarafiga qo'shish.

$$\left(2x+1\right)^{2}-5+5=5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$\left(2x+1\right)^{2}=5$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$2x+1=\sqrt{5} 2x+1=-\sqrt{5}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$2x+1-1=\sqrt{5}-1 2x+1-1=-\sqrt{5}-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$2x=\sqrt{5}-1 2x=-\sqrt{5}-1$$

\sqrt{5} dan 1 ni ayirish.

$$2x=\sqrt{5}-1$$

-\sqrt{5} dan 1 ni ayirish.

$$2x=-\sqrt{5}-1$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x}{2}=\frac{\sqrt{5}-1}{2} \frac{2x}{2}=\frac{-\sqrt{5}-1}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x=\frac{\sqrt{5}-1}{2} x=\frac{-\sqrt{5}-1}{2}$$