x^2+3x+8=0 eching

$$x ^ { 2 } + 3 x + 8 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-3+\sqrt{23}i}{2}\approx -1,5+2,397915762i$$

$$x=\frac{-\sqrt{23}i-3}{2}\approx -1,5-2,397915762i$$

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.

$$x^{2}+3x+8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 3 ni b va 8 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times 8}}{2}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\times 8}}{2}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-32}}{2}$$

9 ni -32 ga qo'shish.

$$x=\frac{-3±\sqrt{-23}}{2}$$

-23 ning kvadrat ildizini chiqarish.

$$x=\frac{-3±\sqrt{23}i}{2}$$

x=\frac{-3±\sqrt{23}i}{2} tenglamasini yeching, bunda ± musbat. -3 ni i\sqrt{23} ga qo'shish.

$$x=\frac{-3+\sqrt{23}i}{2}$$

x=\frac{-3±\sqrt{23}i}{2} tenglamasini yeching, bunda ± manfiy. -3 dan i\sqrt{23} ni ayirish.

$$x=\frac{-\sqrt{23}i-3}{2}$$

Tenglama yechildi.

$$x=\frac{-3+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i-3}{2}$$

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.

$$x^{2}+3x+8=0$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$x^{2}+3x+8-8=-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$x^{2}+3x=-8$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-8+\left(\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{2} kvadratini chiqarish.

$$x^{2}+3x+\frac{9}{4}=-8+\frac{9}{4}$$

-8 ni \frac{9}{4} ga qo'shish.

$$x^{2}+3x+\frac{9}{4}=-\frac{23}{4}$$

x^{2}+3x+\frac{9}{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.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{23}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{23}{4}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{\sqrt{23}i}{2} x+\frac{3}{2}=-\frac{\sqrt{23}i}{2}$$

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

$$x=\frac{-3+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i-3}{2}$$