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x^{2}-6x-4=0
Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.
x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-4\right)}}{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.
x=\frac{-\left(-6\right)±\sqrt{36-4\left(-4\right)}}{2}
-6 kvadratini chiqarish.
x=\frac{-\left(-6\right)±\sqrt{36+16}}{2}
-4 ni -4 marotabaga ko'paytirish.
x=\frac{-\left(-6\right)±\sqrt{52}}{2}
36 ni 16 ga qo'shish.
x=\frac{-\left(-6\right)±2\sqrt{13}}{2}
52 ning kvadrat ildizini chiqarish.
x=\frac{6±2\sqrt{13}}{2}
-6 ning teskarisi 6 ga teng.
x=\frac{2\sqrt{13}+6}{2}
x=\frac{6±2\sqrt{13}}{2} tenglamasini yeching, bunda ± musbat. 6 ni 2\sqrt{13} ga qo'shish.
x=\sqrt{13}+3
6+2\sqrt{13} ni 2 ga bo'lish.
x=\frac{6-2\sqrt{13}}{2}
x=\frac{6±2\sqrt{13}}{2} tenglamasini yeching, bunda ± manfiy. 6 dan 2\sqrt{13} ni ayirish.
x=3-\sqrt{13}
6-2\sqrt{13} ni 2 ga bo'lish.
x^{2}-6x-4=\left(x-\left(\sqrt{13}+3\right)\right)\left(x-\left(3-\sqrt{13}\right)\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 3+\sqrt{13} ga va x_{2} uchun 3-\sqrt{13} ga bo‘ling.