x^2+2x+1=5 eching

$$x ^ { 2 } + 2 x + 1 = 5$$

x uchun yechish (complex solution)

$$x=\sqrt{5}-1\approx 1,236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3,236067977$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x^{2}+2x+1=5$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x^{2}+2x+1-5=5-5$$

O‘zidan 5 ayirilsa 0 qoladi.

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

1 dan 5 ni ayirish.

$$x^{2}+2x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 2 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-4\right)}}{2}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\left(-4\right)}}{2}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

4 ni 16 ga qo'shish.

$$x=\frac{-2±\sqrt{20}}{2}$$

20 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{5}}{2}$$

x=\frac{-2±2\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{5} ga qo'shish.

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

-2+2\sqrt{5} ni 2 ga bo'lish.

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

x=\frac{-2±2\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{5} ni ayirish.

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

-2-2\sqrt{5} ni 2 ga bo'lish.

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan 1 ni ayirish.

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

x uchun yechish

$$x=\sqrt{5}-1\approx 1,236067977$$

$$x=-\sqrt{5}-1\approx -3,236067977$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x^{2}+2x+1=5$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x^{2}+2x+1-5=5-5$$

O‘zidan 5 ayirilsa 0 qoladi.

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

1 dan 5 ni ayirish.

$$x^{2}+2x-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 2 ni b va -4 ni c bilan almashtiring.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-4\right)}}{2}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\left(-4\right)}}{2}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

4 ni 16 ga qo'shish.

$$x=\frac{-2±\sqrt{20}}{2}$$

20 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{5}}{2}$$

x=\frac{-2±2\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{5} ga qo'shish.

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

-2+2\sqrt{5} ni 2 ga bo'lish.

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

x=\frac{-2±2\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{5} ni ayirish.

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

-2-2\sqrt{5} ni 2 ga bo'lish.

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan 1 ni ayirish.

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