5x^2+5x+9=0 eching

$$5 { x }^{ 2 } +5x+9=0$$

x uchun yechish (complex solution)

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2}\approx -0,5+1,24498996i$$

$$x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}\approx -0,5-1,24498996i$$

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.

$$5x^{2}+5x+9=0$$

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 9 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}-4\times 5\times 9}}{2\times 5}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 5\times 9}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-20\times 9}}{2\times 5}$$

-20 ni 9 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-180}}{2\times 5}$$

25 ni -180 ga qo'shish.

$$x=\frac{-5±\sqrt{-155}}{2\times 5}$$

-155 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±\sqrt{155}i}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{155}i}{10}$$

x=\frac{-5±\sqrt{155}i}{10} tenglamasini yeching, bunda ± musbat. -5 ni i\sqrt{155} ga qo'shish.

$$x=\frac{-5+\sqrt{155}i}{10}$$

-5+i\sqrt{155} ni 10 ga bo'lish.

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2}$$

x=\frac{-5±\sqrt{155}i}{10} tenglamasini yeching, bunda ± manfiy. -5 dan i\sqrt{155} ni ayirish.

$$x=\frac{-\sqrt{155}i-5}{10}$$

-5-i\sqrt{155} ni 10 ga bo'lish.

$$x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2} x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}$$

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.

$$5x^{2}+5x+9=0$$

Tenglamaning ikkala tarafidan 9 ni ayirish.

$$5x^{2}+5x+9-9=-9$$

O‘zidan 9 ayirilsa 0 qoladi.

$$5x^{2}+5x=-9$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+5x}{5}=-\frac{9}{5}$$

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

$$x^{2}+\frac{5}{5}x=-\frac{9}{5}$$

5 ni 5 ga bo'lish.

$$x^{2}+x=-\frac{9}{5}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{9}{5}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=-\frac{9}{5}+\frac{1}{4}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{9}{5} ni \frac{1}{4} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+x+\frac{1}{4}=-\frac{31}{20}$$

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.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{31}{20}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{31}{20}}$$

Qisqartirish.

$$x+\frac{1}{2}=\frac{\sqrt{155}i}{10} x+\frac{1}{2}=-\frac{\sqrt{155}i}{10}$$

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

$$x=\frac{\sqrt{155}i}{10}-\frac{1}{2} x=-\frac{\sqrt{155}i}{10}-\frac{1}{2}$$