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x^{2}+12x+24=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{-12±\sqrt{12^{2}-4\times 24}}{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{-12±\sqrt{144-4\times 24}}{2}
12 kvadratini chiqarish.
x=\frac{-12±\sqrt{144-96}}{2}
-4 ni 24 marotabaga ko'paytirish.
x=\frac{-12±\sqrt{48}}{2}
144 ni -96 ga qo'shish.
x=\frac{-12±4\sqrt{3}}{2}
48 ning kvadrat ildizini chiqarish.
x=\frac{4\sqrt{3}-12}{2}
x=\frac{-12±4\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. -12 ni 4\sqrt{3} ga qo'shish.
x=2\sqrt{3}-6
-12+4\sqrt{3} ni 2 ga bo'lish.
x=\frac{-4\sqrt{3}-12}{2}
x=\frac{-12±4\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. -12 dan 4\sqrt{3} ni ayirish.
x=-2\sqrt{3}-6
-12-4\sqrt{3} ni 2 ga bo'lish.
x^{2}+12x+24=\left(x-\left(2\sqrt{3}-6\right)\right)\left(x-\left(-2\sqrt{3}-6\right)\right)
ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -6+2\sqrt{3} ga va x_{2} uchun -6-2\sqrt{3} ga bo‘ling.