n^2+9n+4 eching

$$n ^ { 2 } + 9 n + 4$$

Omil

$$\left(n-\frac{-\sqrt{65}-9}{2}\right)\left(n-\frac{\sqrt{65}-9}{2}\right)$$

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.

$$n^{2}+9n+4=0$$

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.

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

9 kvadratini chiqarish.

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

-4 ni 4 marotabaga ko'paytirish.

$$n=\frac{-9±\sqrt{81-16}}{2}$$

81 ni -16 ga qo'shish.

$$n=\frac{-9±\sqrt{65}}{2}$$

n=\frac{-9±\sqrt{65}}{2} tenglamasini yeching, bunda ± musbat. -9 ni \sqrt{65} ga qo'shish.

$$n=\frac{\sqrt{65}-9}{2}$$

n=\frac{-9±\sqrt{65}}{2} tenglamasini yeching, bunda ± manfiy. -9 dan \sqrt{65} ni ayirish.

$$n=\frac{-\sqrt{65}-9}{2}$$

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{65}}{2} ga va x_{2} uchun \frac{-9-\sqrt{65}}{2} ga bo‘ling.

$$n^{2}+9n+4=\left(n-\frac{\sqrt{65}-9}{2}\right)\left(n-\frac{-\sqrt{65}-9}{2}\right)$$

Baholash

$$n^{2}+9n+4$$