2y+y^2-6=0 eching

$$2 y + y ^ { 2 } - 6 = 0$$

y uchun yechish (complex solution)

$$y=\sqrt{7}-1\approx 1,645751311$$

$$y=-\left(\sqrt{7}+1\right)\approx -3,645751311$$

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.

$$y^{2}+2y-6=0$$

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 -6 ni c bilan almashtiring.

$$y=\frac{-2±\sqrt{2^{2}-4\left(-6\right)}}{2}$$

2 kvadratini chiqarish.

$$y=\frac{-2±\sqrt{4-4\left(-6\right)}}{2}$$

-4 ni -6 marotabaga ko'paytirish.

$$y=\frac{-2±\sqrt{4+24}}{2}$$

4 ni 24 ga qo'shish.

$$y=\frac{-2±\sqrt{28}}{2}$$

28 ning kvadrat ildizini chiqarish.

$$y=\frac{-2±2\sqrt{7}}{2}$$

y=\frac{-2±2\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{7} ga qo'shish.

$$y=\frac{2\sqrt{7}-2}{2}$$

-2+2\sqrt{7} ni 2 ga bo'lish.

$$y=\sqrt{7}-1$$

y=\frac{-2±2\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{7} ni ayirish.

$$y=\frac{-2\sqrt{7}-2}{2}$$

-2-2\sqrt{7} ni 2 ga bo'lish.

$$y=-\sqrt{7}-1$$

Tenglama yechildi.

$$y=\sqrt{7}-1 y=-\sqrt{7}-1$$

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.

$$y^{2}+2y-6=0$$

6 ni tenglamaning ikkala tarafiga qo'shish.

$$y^{2}+2y-6-\left(-6\right)=-\left(-6\right)$$

O‘zidan -6 ayirilsa 0 qoladi.

$$y^{2}+2y=-\left(-6\right)$$

0 dan -6 ni ayirish.

$$y^{2}+2y=6$$

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.

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

1 kvadratini chiqarish.

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

6 ni 1 ga qo'shish.

$$y^{2}+2y+1=7$$

y^{2}+2y+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.

$$\left(y+1\right)^{2}=7$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y+1\right)^{2}}=\sqrt{7}$$

Qisqartirish.

$$y+1=\sqrt{7} y+1=-\sqrt{7}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$y=\sqrt{7}-1 y=-\sqrt{7}-1$$

y uchun yechish

$$y=\sqrt{7}-1\approx 1,645751311$$

$$y=-\sqrt{7}-1\approx -3,645751311$$

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.

$$y^{2}+2y-6=0$$

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 -6 ni c bilan almashtiring.

$$y=\frac{-2±\sqrt{2^{2}-4\left(-6\right)}}{2}$$

2 kvadratini chiqarish.

$$y=\frac{-2±\sqrt{4-4\left(-6\right)}}{2}$$

-4 ni -6 marotabaga ko'paytirish.

$$y=\frac{-2±\sqrt{4+24}}{2}$$

4 ni 24 ga qo'shish.

$$y=\frac{-2±\sqrt{28}}{2}$$

28 ning kvadrat ildizini chiqarish.

$$y=\frac{-2±2\sqrt{7}}{2}$$

y=\frac{-2±2\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{7} ga qo'shish.

$$y=\frac{2\sqrt{7}-2}{2}$$

-2+2\sqrt{7} ni 2 ga bo'lish.

$$y=\sqrt{7}-1$$

y=\frac{-2±2\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{7} ni ayirish.

$$y=\frac{-2\sqrt{7}-2}{2}$$

-2-2\sqrt{7} ni 2 ga bo'lish.

$$y=-\sqrt{7}-1$$

Tenglama yechildi.

$$y=\sqrt{7}-1 y=-\sqrt{7}-1$$

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.

$$y^{2}+2y-6=0$$

6 ni tenglamaning ikkala tarafiga qo'shish.

$$y^{2}+2y-6-\left(-6\right)=-\left(-6\right)$$

O‘zidan -6 ayirilsa 0 qoladi.

$$y^{2}+2y=-\left(-6\right)$$

0 dan -6 ni ayirish.

$$y^{2}+2y=6$$

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.

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

1 kvadratini chiqarish.

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

6 ni 1 ga qo'shish.

$$y^{2}+2y+1=7$$

y^{2}+2y+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.

$$\left(y+1\right)^{2}=7$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y+1\right)^{2}}=\sqrt{7}$$

Qisqartirish.

$$y+1=\sqrt{7} y+1=-\sqrt{7}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$y=\sqrt{7}-1 y=-\sqrt{7}-1$$