2x^2+8x+9=0 eching

$$2 x ^ { 2 } + 8 x + 9 = 0$$

x uchun yechish (complex solution)

$$x=\frac{\sqrt{2}i}{2}-2\approx -2+0,707106781i$$

$$x=-\frac{\sqrt{2}i}{2}-2\approx -2-0,707106781i$$

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.

$$2x^{2}+8x+9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 8 ni b va 9 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\times 9}}{2\times 2}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\times 2\times 9}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-8\times 9}}{2\times 2}$$

-8 ni 9 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-72}}{2\times 2}$$

64 ni -72 ga qo'shish.

$$x=\frac{-8±\sqrt{-8}}{2\times 2}$$

-8 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{2}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-8±2\sqrt{2}i}{4}$$

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

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

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

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

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

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

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

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

Tenglama yechildi.

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

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.

$$2x^{2}+8x+9=0$$

Tenglamaning ikkala tarafidan 9 ni ayirish.

$$2x^{2}+8x+9-9=-9$$

O‘zidan 9 ayirilsa 0 qoladi.

$$2x^{2}+8x=-9$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+8x}{2}=-\frac{9}{2}$$

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

$$x^{2}+\frac{8}{2}x=-\frac{9}{2}$$

8 ni 2 ga bo'lish.

$$x^{2}+4x=-\frac{9}{2}$$

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.

$$x^{2}+4x+2^{2}=-\frac{9}{2}+2^{2}$$

2 kvadratini chiqarish.

$$x^{2}+4x+4=-\frac{9}{2}+4$$

-\frac{9}{2} ni 4 ga qo'shish.

$$x^{2}+4x+4=-\frac{1}{2}$$

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.

$$\left(x+2\right)^{2}=-\frac{1}{2}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-\frac{1}{2}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan 2 ni ayirish.

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