f(x)=x^2+6x-1 eching

$$f ( x ) = x ^ { 2 } + 6 x - 1$$

Omil

$$\left(x-\left(-\sqrt{10}-3\right)\right)\left(x-\left(\sqrt{10}-3\right)\right)$$

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.

$$x^{2}+6x-1=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)}}{2}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+4}}{2}$$

36 ni 4 ga qo'shish.

$$x=\frac{-6±\sqrt{40}}{2}$$

40 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{10}}{2}$$

x=\frac{-6±2\sqrt{10}}{2} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{10} ga qo'shish.

$$x=\frac{2\sqrt{10}-6}{2}$$

-6+2\sqrt{10} ni 2 ga bo'lish.

$$x=\sqrt{10}-3$$

x=\frac{-6±2\sqrt{10}}{2} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{10} ni ayirish.

$$x=\frac{-2\sqrt{10}-6}{2}$$

-6-2\sqrt{10} ni 2 ga bo'lish.

$$x=-\sqrt{10}-3$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -3+\sqrt{10} ga va x_{2} uchun -3-\sqrt{10} ga bo‘ling.

$$x^{2}+6x-1=\left(x-\left(\sqrt{10}-3\right)\right)\left(x-\left(-\sqrt{10}-3\right)\right)$$

Baholash

$$x^{2}+6x-1$$