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2x^{2}-x=\frac{1}{2}
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.
2x^{2}-x-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}
Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.
2x^{2}-x-\frac{1}{2}=0
O‘zidan \frac{1}{2} ayirilsa 0 qoladi.
x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-\frac{1}{2}\right)}}{2\times 2}
Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -1 ni b va -\frac{1}{2} ni c bilan almashtiring.
x=\frac{-\left(-1\right)±\sqrt{1-8\left(-\frac{1}{2}\right)}}{2\times 2}
-4 ni 2 marotabaga ko'paytirish.
x=\frac{-\left(-1\right)±\sqrt{1+4}}{2\times 2}
-8 ni -\frac{1}{2} marotabaga ko'paytirish.
x=\frac{-\left(-1\right)±\sqrt{5}}{2\times 2}
1 ni 4 ga qo'shish.
x=\frac{1±\sqrt{5}}{2\times 2}
-1 ning teskarisi 1 ga teng.
x=\frac{1±\sqrt{5}}{4}
2 ni 2 marotabaga ko'paytirish.
x=\frac{\sqrt{5}+1}{4}
x=\frac{1±\sqrt{5}}{4} tenglamasini yeching, bunda ± musbat. 1 ni \sqrt{5} ga qo'shish.
x=\frac{1-\sqrt{5}}{4}
x=\frac{1±\sqrt{5}}{4} tenglamasini yeching, bunda ± manfiy. 1 dan \sqrt{5} ni ayirish.
x=\frac{\sqrt{5}+1}{4} x=\frac{1-\sqrt{5}}{4}
Tenglama yechildi.
2x^{2}-x=\frac{1}{2}
Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.
\frac{2x^{2}-x}{2}=\frac{\frac{1}{2}}{2}
Ikki tarafini 2 ga bo‘ling.
x^{2}-\frac{1}{2}x=\frac{\frac{1}{2}}{2}
2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.
x^{2}-\frac{1}{2}x=\frac{1}{4}
\frac{1}{2} ni 2 ga bo'lish.
x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1}{4}+\left(-\frac{1}{4}\right)^{2}
-\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{4} olish uchun. Keyin, -\frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.
x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{4}+\frac{1}{16}
Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{4} kvadratini chiqarish.
x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{5}{16}
Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{4} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.
\left(x-\frac{1}{4}\right)^{2}=\frac{5}{16}
x^{2}-\frac{1}{2}x+\frac{1}{16} 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{1}{4}\right)^{2}}=\sqrt{\frac{5}{16}}
Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.
x-\frac{1}{4}=\frac{\sqrt{5}}{4} x-\frac{1}{4}=-\frac{\sqrt{5}}{4}
Qisqartirish.
x=\frac{\sqrt{5}+1}{4} x=\frac{1-\sqrt{5}}{4}
\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.