b^2+2b-5=0 eching

$$b ^ { 2 } + 2 b - 5 = 0$$

b uchun yechish (complex solution)

$$b=\sqrt{6}-1\approx 1,449489743$$

$$b=-\left(\sqrt{6}+1\right)\approx -3,449489743$$

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.

$$b^{2}+2b-5=0$$

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 -5 ni c bilan almashtiring.

$$b=\frac{-2±\sqrt{2^{2}-4\left(-5\right)}}{2}$$

2 kvadratini chiqarish.

$$b=\frac{-2±\sqrt{4-4\left(-5\right)}}{2}$$

-4 ni -5 marotabaga ko'paytirish.

$$b=\frac{-2±\sqrt{4+20}}{2}$$

4 ni 20 ga qo'shish.

$$b=\frac{-2±\sqrt{24}}{2}$$

24 ning kvadrat ildizini chiqarish.

$$b=\frac{-2±2\sqrt{6}}{2}$$

b=\frac{-2±2\sqrt{6}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{6} ga qo'shish.

$$b=\frac{2\sqrt{6}-2}{2}$$

-2+2\sqrt{6} ni 2 ga bo'lish.

$$b=\sqrt{6}-1$$

b=\frac{-2±2\sqrt{6}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{6} ni ayirish.

$$b=\frac{-2\sqrt{6}-2}{2}$$

-2-2\sqrt{6} ni 2 ga bo'lish.

$$b=-\sqrt{6}-1$$

Tenglama yechildi.

$$b=\sqrt{6}-1 b=-\sqrt{6}-1$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$b^{2}+2b-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -5 ayirilsa 0 qoladi.

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

0 dan -5 ni ayirish.

$$b^{2}+2b=5$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

1 kvadratini chiqarish.

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

5 ni 1 ga qo'shish.

$$b^{2}+2b+1=6$$

b^{2}+2b+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.

$$\left(b+1\right)^{2}=6$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(b+1\right)^{2}}=\sqrt{6}$$

Qisqartirish.

$$b+1=\sqrt{6} b+1=-\sqrt{6}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$b=\sqrt{6}-1 b=-\sqrt{6}-1$$

b uchun yechish

$$b=\sqrt{6}-1\approx 1,449489743$$

$$b=-\sqrt{6}-1\approx -3,449489743$$

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.

$$b^{2}+2b-5=0$$

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 -5 ni c bilan almashtiring.

$$b=\frac{-2±\sqrt{2^{2}-4\left(-5\right)}}{2}$$

2 kvadratini chiqarish.

$$b=\frac{-2±\sqrt{4-4\left(-5\right)}}{2}$$

-4 ni -5 marotabaga ko'paytirish.

$$b=\frac{-2±\sqrt{4+20}}{2}$$

4 ni 20 ga qo'shish.

$$b=\frac{-2±\sqrt{24}}{2}$$

24 ning kvadrat ildizini chiqarish.

$$b=\frac{-2±2\sqrt{6}}{2}$$

b=\frac{-2±2\sqrt{6}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{6} ga qo'shish.

$$b=\frac{2\sqrt{6}-2}{2}$$

-2+2\sqrt{6} ni 2 ga bo'lish.

$$b=\sqrt{6}-1$$

b=\frac{-2±2\sqrt{6}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{6} ni ayirish.

$$b=\frac{-2\sqrt{6}-2}{2}$$

-2-2\sqrt{6} ni 2 ga bo'lish.

$$b=-\sqrt{6}-1$$

Tenglama yechildi.

$$b=\sqrt{6}-1 b=-\sqrt{6}-1$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$b^{2}+2b-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -5 ayirilsa 0 qoladi.

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

0 dan -5 ni ayirish.

$$b^{2}+2b=5$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

1 kvadratini chiqarish.

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

5 ni 1 ga qo'shish.

$$b^{2}+2b+1=6$$

b^{2}+2b+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.

$$\left(b+1\right)^{2}=6$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(b+1\right)^{2}}=\sqrt{6}$$

Qisqartirish.

$$b+1=\sqrt{6} b+1=-\sqrt{6}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$b=\sqrt{6}-1 b=-\sqrt{6}-1$$