2x^2+12x+40=0 eching

$$2x^2+12x+40=0$$

x uchun yechish (complex solution)

$$x=-3+\sqrt{11}i\approx -3+3,31662479i$$

$$x=-\sqrt{11}i-3\approx -3-3,31662479i$$

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.

$$2x^{2}+12x+40=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 12 ni b va 40 ni c bilan almashtiring.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2\times 40}}{2\times 2}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 2\times 40}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-8\times 40}}{2\times 2}$$

-8 ni 40 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-320}}{2\times 2}$$

144 ni -320 ga qo'shish.

$$x=\frac{-12±\sqrt{-176}}{2\times 2}$$

-176 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±4\sqrt{11}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-12±4\sqrt{11}i}{4}$$

x=\frac{-12±4\sqrt{11}i}{4} tenglamasini yeching, bunda ± musbat. -12 ni 4i\sqrt{11} ga qo'shish.

$$x=\frac{-12+4\sqrt{11}i}{4}$$

-12+4i\sqrt{11} ni 4 ga bo'lish.

$$x=-3+\sqrt{11}i$$

x=\frac{-12±4\sqrt{11}i}{4} tenglamasini yeching, bunda ± manfiy. -12 dan 4i\sqrt{11} ni ayirish.

$$x=\frac{-4\sqrt{11}i-12}{4}$$

-12-4i\sqrt{11} ni 4 ga bo'lish.

$$x=-\sqrt{11}i-3$$

Tenglama yechildi.

$$x=-3+\sqrt{11}i x=-\sqrt{11}i-3$$

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.

$$2x^{2}+12x+40=0$$

Tenglamaning ikkala tarafidan 40 ni ayirish.

$$2x^{2}+12x+40-40=-40$$

O‘zidan 40 ayirilsa 0 qoladi.

$$2x^{2}+12x=-40$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+12x}{2}=-\frac{40}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{12}{2}x=-\frac{40}{2}$$

12 ni 2 ga bo'lish.

$$x^{2}+6x=-\frac{40}{2}$$

-40 ni 2 ga bo'lish.

$$x^{2}+6x=-20$$

6 ni bo‘lish, x shartining koeffitsienti, 2 ga 3 olish uchun. Keyin, 3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+6x+3^{2}=-20+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=-20+9$$

-20 ni 9 ga qo'shish.

$$x^{2}+6x+9=-11$$

x^{2}+6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+3\right)^{2}=-11$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{-11}$$

Qisqartirish.

$$x+3=\sqrt{11}i x+3=-\sqrt{11}i$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=-3+\sqrt{11}i x=-\sqrt{11}i-3$$