f(x)=x^2-5x+1 eching

$$f ( x ) = x ^ { 2 } - 5 x + 1$$

Omil

$$\left(x-\frac{5-\sqrt{21}}{2}\right)\left(x-\frac{\sqrt{21}+5}{2}\right)$$

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.

$$x^{2}-5x+1=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4}}{2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4}}{2}$$

25 ni -4 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{21}}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{21}}{2}$$

x=\frac{5±\sqrt{21}}{2} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{21} ga qo'shish.

$$x=\frac{\sqrt{21}+5}{2}$$

x=\frac{5±\sqrt{21}}{2} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{21} ni ayirish.

$$x=\frac{5-\sqrt{21}}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{5+\sqrt{21}}{2} ga va x_{2} uchun \frac{5-\sqrt{21}}{2} ga bo‘ling.

$$x^{2}-5x+1=\left(x-\frac{\sqrt{21}+5}{2}\right)\left(x-\frac{5-\sqrt{21}}{2}\right)$$

Baholash

$$x^{2}-5x+1$$