x^2+sqrt{6}x+5=0 eching

$$x ^ { 2 } + \sqrt { 6 } x + 5 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-\sqrt{6}+\sqrt{14}i}{2}\approx -1,224744871+1,870828693i$$

$$x=\frac{-\sqrt{14}i-\sqrt{6}}{2}\approx -1,224744871-1,870828693i$$

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.

$$x^{2}+\sqrt{6}x+5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, \sqrt{6} ni b va 5 ni c bilan almashtiring.

$$x=\frac{-\sqrt{6}±\sqrt{\left(\sqrt{6}\right)^{2}-4\times 5}}{2}$$

\sqrt{6} kvadratini chiqarish.

$$x=\frac{-\sqrt{6}±\sqrt{6-4\times 5}}{2}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\sqrt{6}±\sqrt{6-20}}{2}$$

6 ni -20 ga qo'shish.

$$x=\frac{-\sqrt{6}±\sqrt{-14}}{2}$$

-14 ning kvadrat ildizini chiqarish.

$$x=\frac{-\sqrt{6}±\sqrt{14}i}{2}$$

x=\frac{-\sqrt{6}±\sqrt{14}i}{2} tenglamasini yeching, bunda ± musbat. -\sqrt{6} ni i\sqrt{14} ga qo'shish.

$$x=\frac{-\sqrt{6}+\sqrt{14}i}{2}$$

x=\frac{-\sqrt{6}±\sqrt{14}i}{2} tenglamasini yeching, bunda ± manfiy. -\sqrt{6} dan i\sqrt{14} ni ayirish.

$$x=\frac{-\sqrt{14}i-\sqrt{6}}{2}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{6}+\sqrt{14}i}{2} x=\frac{-\sqrt{14}i-\sqrt{6}}{2}$$

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.

$$x^{2}+\sqrt{6}x+5=0$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x^{2}+\sqrt{6}x+5-5=-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$x^{2}+\sqrt{6}x=-5$$

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

$$x^{2}+\sqrt{6}x+\left(\frac{\sqrt{6}}{2}\right)^{2}=-5+\left(\frac{\sqrt{6}}{2}\right)^{2}$$

\frac{\sqrt{6}}{2} kvadratini chiqarish.

$$x^{2}+\sqrt{6}x+\frac{3}{2}=-5+\frac{3}{2}$$

-5 ni \frac{3}{2} ga qo'shish.

$$x^{2}+\sqrt{6}x+\frac{3}{2}=-\frac{7}{2}$$

x^{2}+\sqrt{6}x+\frac{3}{2} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{\sqrt{6}}{2}\right)^{2}=-\frac{7}{2}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{\sqrt{6}}{2}\right)^{2}}=\sqrt{-\frac{7}{2}}$$

Qisqartirish.

$$x+\frac{\sqrt{6}}{2}=\frac{\sqrt{14}i}{2} x+\frac{\sqrt{6}}{2}=-\frac{\sqrt{14}i}{2}$$

Tenglamaning ikkala tarafidan \frac{\sqrt{6}}{2} ni ayirish.

$$x=\frac{-\sqrt{6}+\sqrt{14}i}{2} x=\frac{-\sqrt{14}i-\sqrt{6}}{2}$$