1x^2+5x+1 eching

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

Omil

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

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.

$$x=\frac{-5±\sqrt{5^{2}-4}}{2}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4}}{2}$$

25 ni -4 ga qo'shish.

$$x=\frac{-5±\sqrt{21}}{2}$$

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.

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

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.

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

Baholash

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