5x^2+5x+5=0 eching

$$5 x ^ { 2 } + 5 x + 5 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-1+\sqrt{3}i}{2}\approx -0,5+0,866025404i$$

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0,5-0,866025404i$$

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.

$$5x^{2}+5x+5=0$$

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

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

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 5\times 5}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-20\times 5}}{2\times 5}$$

-20 ni 5 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-100}}{2\times 5}$$

25 ni -100 ga qo'shish.

$$x=\frac{-5±\sqrt{-75}}{2\times 5}$$

-75 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±5\sqrt{3}i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-5±5\sqrt{3}i}{10}$$

x=\frac{-5±5\sqrt{3}i}{10} tenglamasini yeching, bunda ± musbat. -5 ni 5i\sqrt{3} ga qo'shish.

$$x=\frac{-5+5\sqrt{3}i}{10}$$

-5+5i\sqrt{3} ni 10 ga bo'lish.

$$x=\frac{-1+\sqrt{3}i}{2}$$

x=\frac{-5±5\sqrt{3}i}{10} tenglamasini yeching, bunda ± manfiy. -5 dan 5i\sqrt{3} ni ayirish.

$$x=\frac{-5\sqrt{3}i-5}{10}$$

-5-5i\sqrt{3} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{3}i-1}{2}$$

Tenglama yechildi.

$$x=\frac{-1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-1}{2}$$

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.

$$5x^{2}+5x+5=0$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$5x^{2}+5x+5-5=-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$5x^{2}+5x=-5$$

Ikki tarafini 5 ga bo‘ling.

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

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

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

5 ni 5 ga bo'lish.

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

-5 ni 5 ga bo'lish.

$$x^{2}+x=-1$$

1 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{2} olish uchun. Keyin, \frac{1}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

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

-1 ni \frac{1}{4} ga qo'shish.

$$x^{2}+x+\frac{1}{4}=-\frac{3}{4}$$

x^{2}+x+\frac{1}{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.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{3}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{3}{4}}$$

Qisqartirish.

$$x+\frac{1}{2}=\frac{\sqrt{3}i}{2} x+\frac{1}{2}=-\frac{\sqrt{3}i}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$x=\frac{-1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-1}{2}$$