x^2+12x-9 eching

$${ x }^{ 2 } +12x-9$$

Omil

$$\left(x-\left(-3\sqrt{5}-6\right)\right)\left(x-\left(3\sqrt{5}-6\right)\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$x^{2}+12x-9=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-9\right)}}{2}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\left(-9\right)}}{2}$$

-4 ni -9 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+36}}{2}$$

144 ni 36 ga qo'shish.

$$x=\frac{-12±\sqrt{180}}{2}$$

180 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±6\sqrt{5}}{2}$$

x=\frac{-12±6\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -12 ni 6\sqrt{5} ga qo'shish.

$$x=\frac{6\sqrt{5}-12}{2}$$

-12+6\sqrt{5} ni 2 ga bo'lish.

$$x=3\sqrt{5}-6$$

x=\frac{-12±6\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -12 dan 6\sqrt{5} ni ayirish.

$$x=\frac{-6\sqrt{5}-12}{2}$$

-12-6\sqrt{5} ni 2 ga bo'lish.

$$x=-3\sqrt{5}-6$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -6+3\sqrt{5} ga va x_{2} uchun -6-3\sqrt{5} ga bo‘ling.

$$x^{2}+12x-9=\left(x-\left(3\sqrt{5}-6\right)\right)\left(x-\left(-3\sqrt{5}-6\right)\right)$$

Baholash

$$x^{2}+12x-9$$