p(x)=x^2-6x+4 eching

$$p ( x ) = x ^ { 2 } - 6 x + 4$$

Omil

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

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.

$$x^{2}-6x+4=0$$

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.

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

-6 kvadratini chiqarish.

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

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36-16}}{2}$$

36 ni -16 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{20}}{2}$$

20 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-6\right)±2\sqrt{5}}{2}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±2\sqrt{5}}{2}$$

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

$$x=\frac{2\sqrt{5}+6}{2}$$

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

$$x=\sqrt{5}+3$$

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

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

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

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

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

$$x^{2}-6x+4=\left(x-\left(\sqrt{5}+3\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)$$

Baholash

$$x^{2}-6x+4$$