x^2+4x-16=0 eching

$$x ^ { 2 } + 4 x - 16 = 0$$

x uchun yechish

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

$$x=-2\sqrt{5}-2\approx -6,472135955$$

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.

$$x^{2}+4x-16=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 4 ni b va -16 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-16\right)}}{2}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\left(-16\right)}}{2}$$

-4 ni -16 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16+64}}{2}$$

16 ni 64 ga qo'shish.

$$x=\frac{-4±\sqrt{80}}{2}$$

80 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4\sqrt{5}}{2}$$

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

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

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

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

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

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

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

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

Tenglama yechildi.

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

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.

$$x^{2}+4x-16=0$$

16 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+4x-16-\left(-16\right)=-\left(-16\right)$$

O‘zidan -16 ayirilsa 0 qoladi.

$$x^{2}+4x=-\left(-16\right)$$

0 dan -16 ni ayirish.

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

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

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

2 kvadratini chiqarish.

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

16 ni 4 ga qo'shish.

$$x^{2}+4x+4=20$$

x^{2}+4x+4 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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