5x^2+9x+1 eching

$$5 { x }^{ 2 } +9x+1$$

Omil

$$5\left(x-\frac{-\sqrt{61}-9}{10}\right)\left(x-\frac{\sqrt{61}-9}{10}\right)$$

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.

$$5x^{2}+9x+1=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\times 5}}{2\times 5}$$

9 kvadratini chiqarish.

$$x=\frac{-9±\sqrt{81-4\times 5}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81-20}}{2\times 5}$$

81 ni -20 ga qo'shish.

$$x=\frac{-9±\sqrt{61}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{61}}{10}$$

x=\frac{-9±\sqrt{61}}{10} tenglamasini yeching, bunda ± musbat. -9 ni \sqrt{61} ga qo'shish.

$$x=\frac{\sqrt{61}-9}{10}$$

x=\frac{-9±\sqrt{61}}{10} tenglamasini yeching, bunda ± manfiy. -9 dan \sqrt{61} ni ayirish.

$$x=\frac{-\sqrt{61}-9}{10}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-9+\sqrt{61}}{10} ga va x_{2} uchun \frac{-9-\sqrt{61}}{10} ga bo‘ling.

$$5x^{2}+9x+1=5\left(x-\frac{\sqrt{61}-9}{10}\right)\left(x-\frac{-\sqrt{61}-9}{10}\right)$$

Baholash

$$5x^{2}+9x+1$$