x^2+4x+36=0 eching

$$x ^ { 2 } + 4 x + 36 = 0$$

x uchun yechish (complex solution)

$$x=-2+4\sqrt{2}i\approx -2+5,656854249i$$

$$x=-4\sqrt{2}i-2\approx -2-5,656854249i$$

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.

$$x^{2}+4x+36=0$$

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 36 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 36}}{2}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 36}}{2}$$

-4 ni 36 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-144}}{2}$$

16 ni -144 ga qo'shish.

$$x=\frac{-4±\sqrt{-128}}{2}$$

-128 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±8\sqrt{2}i}{2}$$

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

$$x=\frac{-4+2\times 2^{\frac{5}{2}}i}{2}$$

-4+2i\times 2^{\frac{5}{2}} ni 2 ga bo'lish.

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

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

$$x=\frac{-2\times 2^{\frac{5}{2}}i-4}{2}$$

-4-2i\times 2^{\frac{5}{2}} ni 2 ga bo'lish.

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

Tenglama yechildi.

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

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.

$$x^{2}+4x+36=0$$

Tenglamaning ikkala tarafidan 36 ni ayirish.

$$x^{2}+4x+36-36=-36$$

O‘zidan 36 ayirilsa 0 qoladi.

$$x^{2}+4x=-36$$

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.

$$x^{2}+4x+2^{2}=-36+2^{2}$$

2 kvadratini chiqarish.

$$x^{2}+4x+4=-36+4$$

-36 ni 4 ga qo'shish.

$$x^{2}+4x+4=-32$$

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.

$$\left(x+2\right)^{2}=-32$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-32}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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