Asosiy tarkibga oʻtish
x uchun yechish (complex solution)
Tick mark Image
Grafik

Veb-qidiruvdagi o'xshash muammolar

Baham ko'rish

5x^{2}+3x+1=0
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=\frac{-3±\sqrt{3^{2}-4\times 5}}{2\times 5}
Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 3 ni b va 1 ni c bilan almashtiring.
x=\frac{-3±\sqrt{9-4\times 5}}{2\times 5}
3 kvadratini chiqarish.
x=\frac{-3±\sqrt{9-20}}{2\times 5}
-4 ni 5 marotabaga ko'paytirish.
x=\frac{-3±\sqrt{-11}}{2\times 5}
9 ni -20 ga qo'shish.
x=\frac{-3±\sqrt{11}i}{2\times 5}
-11 ning kvadrat ildizini chiqarish.
x=\frac{-3±\sqrt{11}i}{10}
2 ni 5 marotabaga ko'paytirish.
x=\frac{-3+\sqrt{11}i}{10}
x=\frac{-3±\sqrt{11}i}{10} tenglamasini yeching, bunda ± musbat. -3 ni i\sqrt{11} ga qo'shish.
x=\frac{-\sqrt{11}i-3}{10}
x=\frac{-3±\sqrt{11}i}{10} tenglamasini yeching, bunda ± manfiy. -3 dan i\sqrt{11} ni ayirish.
x=\frac{-3+\sqrt{11}i}{10} x=\frac{-\sqrt{11}i-3}{10}
Tenglama yechildi.
5x^{2}+3x+1=0
Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.
5x^{2}+3x+1-1=-1
Tenglamaning ikkala tarafidan 1 ni ayirish.
5x^{2}+3x=-1
O‘zidan 1 ayirilsa 0 qoladi.
\frac{5x^{2}+3x}{5}=-\frac{1}{5}
Ikki tarafini 5 ga bo‘ling.
x^{2}+\frac{3}{5}x=-\frac{1}{5}
5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.
x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=-\frac{1}{5}+\left(\frac{3}{10}\right)^{2}
\frac{3}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{10} olish uchun. Keyin, \frac{3}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.
x^{2}+\frac{3}{5}x+\frac{9}{100}=-\frac{1}{5}+\frac{9}{100}
Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{10} kvadratini chiqarish.
x^{2}+\frac{3}{5}x+\frac{9}{100}=-\frac{11}{100}
Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{5} ni \frac{9}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.
\left(x+\frac{3}{10}\right)^{2}=-\frac{11}{100}
x^{2}+\frac{3}{5}x+\frac{9}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.
\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{-\frac{11}{100}}
Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.
x+\frac{3}{10}=\frac{\sqrt{11}i}{10} x+\frac{3}{10}=-\frac{\sqrt{11}i}{10}
Qisqartirish.
x=\frac{-3+\sqrt{11}i}{10} x=\frac{-\sqrt{11}i-3}{10}
Tenglamaning ikkala tarafidan \frac{3}{10} ni ayirish.