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5x^{2}-4x+5=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{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\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, -4 ni b va 5 ni c bilan almashtiring.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\times 5}}{2\times 5}
-4 kvadratini chiqarish.
x=\frac{-\left(-4\right)±\sqrt{16-20\times 5}}{2\times 5}
-4 ni 5 marotabaga ko'paytirish.
x=\frac{-\left(-4\right)±\sqrt{16-100}}{2\times 5}
-20 ni 5 marotabaga ko'paytirish.
x=\frac{-\left(-4\right)±\sqrt{-84}}{2\times 5}
16 ni -100 ga qo'shish.
x=\frac{-\left(-4\right)±2\sqrt{21}i}{2\times 5}
-84 ning kvadrat ildizini chiqarish.
x=\frac{4±2\sqrt{21}i}{2\times 5}
-4 ning teskarisi 4 ga teng.
x=\frac{4±2\sqrt{21}i}{10}
2 ni 5 marotabaga ko'paytirish.
x=\frac{4+2\sqrt{21}i}{10}
x=\frac{4±2\sqrt{21}i}{10} tenglamasini yeching, bunda ± musbat. 4 ni 2i\sqrt{21} ga qo'shish.
x=\frac{2+\sqrt{21}i}{5}
4+2i\sqrt{21} ni 10 ga bo'lish.
x=\frac{-2\sqrt{21}i+4}{10}
x=\frac{4±2\sqrt{21}i}{10} tenglamasini yeching, bunda ± manfiy. 4 dan 2i\sqrt{21} ni ayirish.
x=\frac{-\sqrt{21}i+2}{5}
4-2i\sqrt{21} ni 10 ga bo'lish.
x=\frac{2+\sqrt{21}i}{5} x=\frac{-\sqrt{21}i+2}{5}
Tenglama yechildi.
5x^{2}-4x+5=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}-4x+5-5=-5
Tenglamaning ikkala tarafidan 5 ni ayirish.
5x^{2}-4x=-5
O‘zidan 5 ayirilsa 0 qoladi.
\frac{5x^{2}-4x}{5}=-\frac{5}{5}
Ikki tarafini 5 ga bo‘ling.
x^{2}-\frac{4}{5}x=-\frac{5}{5}
5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.
x^{2}-\frac{4}{5}x=-1
-5 ni 5 ga bo'lish.
x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=-1+\left(-\frac{2}{5}\right)^{2}
-\frac{4}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{2}{5} olish uchun. Keyin, -\frac{2}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.
x^{2}-\frac{4}{5}x+\frac{4}{25}=-1+\frac{4}{25}
Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{2}{5} kvadratini chiqarish.
x^{2}-\frac{4}{5}x+\frac{4}{25}=-\frac{21}{25}
-1 ni \frac{4}{25} ga qo'shish.
\left(x-\frac{2}{5}\right)^{2}=-\frac{21}{25}
x^{2}-\frac{4}{5}x+\frac{4}{25} 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{2}{5}\right)^{2}}=\sqrt{-\frac{21}{25}}
Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.
x-\frac{2}{5}=\frac{\sqrt{21}i}{5} x-\frac{2}{5}=-\frac{\sqrt{21}i}{5}
Qisqartirish.
x=\frac{2+\sqrt{21}i}{5} x=\frac{-\sqrt{21}i+2}{5}
\frac{2}{5} ni tenglamaning ikkala tarafiga qo'shish.