x^2+8x+4=-10 eching

$${ x }^{ 2 } +8x+4=-10$$

x uchun yechish (complex solution)

$$x=\sqrt{2}-4\approx -2,585786438$$

$$x=-\left(\sqrt{2}+4\right)\approx -5,414213562$$

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.

$$x^{2}+8x+4=-10$$

10 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+8x+4-\left(-10\right)=-10-\left(-10\right)$$

O‘zidan -10 ayirilsa 0 qoladi.

$$x^{2}+8x+4-\left(-10\right)=0$$

4 dan -10 ni ayirish.

$$x^{2}+8x+14=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 8 ni b va 14 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\times 14}}{2}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 14}}{2}$$

-4 ni 14 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-56}}{2}$$

64 ni -56 ga qo'shish.

$$x=\frac{-8±\sqrt{8}}{2}$$

8 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{2}}{2}$$

x=\frac{-8±2\sqrt{2}}{2} tenglamasini yeching, bunda ± musbat. -8 ni 2\sqrt{2} ga qo'shish.

$$x=\frac{2\sqrt{2}-8}{2}$$

2\sqrt{2}-8 ni 2 ga bo'lish.

$$x=\sqrt{2}-4$$

x=\frac{-8±2\sqrt{2}}{2} tenglamasini yeching, bunda ± manfiy. -8 dan 2\sqrt{2} ni ayirish.

$$x=\frac{-2\sqrt{2}-8}{2}$$

-8-2\sqrt{2} ni 2 ga bo'lish.

$$x=-\sqrt{2}-4$$

Tenglama yechildi.

$$x=\sqrt{2}-4 x=-\sqrt{2}-4$$

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.

$$x^{2}+8x+4=-10$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$x^{2}+8x+4-4=-10-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$x^{2}+8x=-10-4$$

-10 dan 4 ni ayirish.

$$x^{2}+8x=-14$$

8 ni bo‘lish, x shartining koeffitsienti, 2 ga 4 olish uchun. Keyin, 4 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+8x+4^{2}=-14+4^{2}$$

4 kvadratini chiqarish.

$$x^{2}+8x+16=-14+16$$

-14 ni 16 ga qo'shish.

$$x^{2}+8x+16=2$$

x^{2}+8x+16 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+4\right)^{2}=2$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{2}$$

Qisqartirish.

$$x+4=\sqrt{2} x+4=-\sqrt{2}$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$x=\sqrt{2}-4 x=-\sqrt{2}-4$$

x uchun yechish

$$x=\sqrt{2}-4\approx -2,585786438$$

$$x=-\sqrt{2}-4\approx -5,414213562$$

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.

$$x^{2}+8x+4=-10$$

10 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+8x+4-\left(-10\right)=-10-\left(-10\right)$$

O‘zidan -10 ayirilsa 0 qoladi.

$$x^{2}+8x+4-\left(-10\right)=0$$

4 dan -10 ni ayirish.

$$x^{2}+8x+14=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 8 ni b va 14 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\times 14}}{2}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 14}}{2}$$

-4 ni 14 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-56}}{2}$$

64 ni -56 ga qo'shish.

$$x=\frac{-8±\sqrt{8}}{2}$$

8 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{2}}{2}$$

x=\frac{-8±2\sqrt{2}}{2} tenglamasini yeching, bunda ± musbat. -8 ni 2\sqrt{2} ga qo'shish.

$$x=\frac{2\sqrt{2}-8}{2}$$

2\sqrt{2}-8 ni 2 ga bo'lish.

$$x=\sqrt{2}-4$$

x=\frac{-8±2\sqrt{2}}{2} tenglamasini yeching, bunda ± manfiy. -8 dan 2\sqrt{2} ni ayirish.

$$x=\frac{-2\sqrt{2}-8}{2}$$

-8-2\sqrt{2} ni 2 ga bo'lish.

$$x=-\sqrt{2}-4$$

Tenglama yechildi.

$$x=\sqrt{2}-4 x=-\sqrt{2}-4$$

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.

$$x^{2}+8x+4=-10$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$x^{2}+8x+4-4=-10-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$x^{2}+8x=-10-4$$

-10 dan 4 ni ayirish.

$$x^{2}+8x=-14$$

8 ni bo‘lish, x shartining koeffitsienti, 2 ga 4 olish uchun. Keyin, 4 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+8x+4^{2}=-14+4^{2}$$

4 kvadratini chiqarish.

$$x^{2}+8x+16=-14+16$$

-14 ni 16 ga qo'shish.

$$x^{2}+8x+16=2$$

x^{2}+8x+16 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+4\right)^{2}=2$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{2}$$

Qisqartirish.

$$x+4=\sqrt{2} x+4=-\sqrt{2}$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$x=\sqrt{2}-4 x=-\sqrt{2}-4$$