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6x^{2}+4x-3=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{-4±\sqrt{4^{2}-4\times 6\left(-3\right)}}{2\times 6}
Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 6 ni a, 4 ni b va -3 ni c bilan almashtiring.
x=\frac{-4±\sqrt{16-4\times 6\left(-3\right)}}{2\times 6}
4 kvadratini chiqarish.
x=\frac{-4±\sqrt{16-24\left(-3\right)}}{2\times 6}
-4 ni 6 marotabaga ko'paytirish.
x=\frac{-4±\sqrt{16+72}}{2\times 6}
-24 ni -3 marotabaga ko'paytirish.
x=\frac{-4±\sqrt{88}}{2\times 6}
16 ni 72 ga qo'shish.
x=\frac{-4±2\sqrt{22}}{2\times 6}
88 ning kvadrat ildizini chiqarish.
x=\frac{-4±2\sqrt{22}}{12}
2 ni 6 marotabaga ko'paytirish.
x=\frac{2\sqrt{22}-4}{12}
x=\frac{-4±2\sqrt{22}}{12} tenglamasini yeching, bunda ± musbat. -4 ni 2\sqrt{22} ga qo'shish.
x=\frac{\sqrt{22}}{6}-\frac{1}{3}
-4+2\sqrt{22} ni 12 ga bo'lish.
x=\frac{-2\sqrt{22}-4}{12}
x=\frac{-4±2\sqrt{22}}{12} tenglamasini yeching, bunda ± manfiy. -4 dan 2\sqrt{22} ni ayirish.
x=-\frac{\sqrt{22}}{6}-\frac{1}{3}
-4-2\sqrt{22} ni 12 ga bo'lish.
x=\frac{\sqrt{22}}{6}-\frac{1}{3} x=-\frac{\sqrt{22}}{6}-\frac{1}{3}
Tenglama yechildi.
6x^{2}+4x-3=0
Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.
6x^{2}+4x-3-\left(-3\right)=-\left(-3\right)
3 ni tenglamaning ikkala tarafiga qo'shish.
6x^{2}+4x=-\left(-3\right)
O‘zidan -3 ayirilsa 0 qoladi.
6x^{2}+4x=3
0 dan -3 ni ayirish.
\frac{6x^{2}+4x}{6}=\frac{3}{6}
Ikki tarafini 6 ga bo‘ling.
x^{2}+\frac{4}{6}x=\frac{3}{6}
6 ga bo'lish 6 ga ko'paytirishni bekor qiladi.
x^{2}+\frac{2}{3}x=\frac{3}{6}
\frac{4}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.
x^{2}+\frac{2}{3}x=\frac{1}{2}
\frac{3}{6} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.
x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{1}{2}+\left(\frac{1}{3}\right)^{2}
\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{3} olish uchun. Keyin, \frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.
x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{1}{2}+\frac{1}{9}
Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{3} kvadratini chiqarish.
x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{11}{18}
Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.
\left(x+\frac{1}{3}\right)^{2}=\frac{11}{18}
x^{2}+\frac{2}{3}x+\frac{1}{9} 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}{3}\right)^{2}}=\sqrt{\frac{11}{18}}
Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.
x+\frac{1}{3}=\frac{\sqrt{22}}{6} x+\frac{1}{3}=-\frac{\sqrt{22}}{6}
Qisqartirish.
x=\frac{\sqrt{22}}{6}-\frac{1}{3} x=-\frac{\sqrt{22}}{6}-\frac{1}{3}
Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.