x^2+8x-4=0 eching

$${ x }^{ 2 } +8x-4 = 0$$

x uchun yechish

$$x=2\sqrt{5}-4\approx 0,472135955$$

$$x=-2\sqrt{5}-4\approx -8,472135955$$

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.

$$x^{2}+8x-4=0$$

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 -4 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-4\right)}}{2}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\left(-4\right)}}{2}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+16}}{2}$$

64 ni 16 ga qo'shish.

$$x=\frac{-8±\sqrt{80}}{2}$$

80 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±4\sqrt{5}}{2}$$

x=\frac{-8±4\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -8 ni 4\sqrt{5} ga qo'shish.

$$x=\frac{4\sqrt{5}-8}{2}$$

-8+4\sqrt{5} ni 2 ga bo'lish.

$$x=2\sqrt{5}-4$$

x=\frac{-8±4\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -8 dan 4\sqrt{5} ni ayirish.

$$x=\frac{-4\sqrt{5}-8}{2}$$

-8-4\sqrt{5} ni 2 ga bo'lish.

$$x=-2\sqrt{5}-4$$

Tenglama yechildi.

$$x=2\sqrt{5}-4 x=-2\sqrt{5}-4$$

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.

$$x^{2}+8x-4=0$$

4 ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -4 ayirilsa 0 qoladi.

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

0 dan -4 ni ayirish.

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

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.

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

4 kvadratini chiqarish.

$$x^{2}+8x+16=4+16$$

4 ni 16 ga qo'shish.

$$x^{2}+8x+16=20$$

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.

$$\left(x+4\right)^{2}=20$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{20}$$

Qisqartirish.

$$x+4=2\sqrt{5} x+4=-2\sqrt{5}$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$x=2\sqrt{5}-4 x=-2\sqrt{5}-4$$