9x^2+7x+1 eching

$$9 x ^ { 2 } + 7 x + 1$$

Omil

$$9\left(x-\frac{-\sqrt{13}-7}{18}\right)\left(x-\frac{\sqrt{13}-7}{18}\right)$$

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.

$$9x^{2}+7x+1=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 9}}{2\times 9}$$

7 kvadratini chiqarish.

$$x=\frac{-7±\sqrt{49-4\times 9}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49-36}}{2\times 9}$$

49 ni -36 ga qo'shish.

$$x=\frac{-7±\sqrt{13}}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{13}}{18}$$

x=\frac{-7±\sqrt{13}}{18} tenglamasini yeching, bunda ± musbat. -7 ni \sqrt{13} ga qo'shish.

$$x=\frac{\sqrt{13}-7}{18}$$

x=\frac{-7±\sqrt{13}}{18} tenglamasini yeching, bunda ± manfiy. -7 dan \sqrt{13} ni ayirish.

$$x=\frac{-\sqrt{13}-7}{18}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-7+\sqrt{13}}{18} ga va x_{2} uchun \frac{-7-\sqrt{13}}{18} ga bo‘ling.

$$9x^{2}+7x+1=9\left(x-\frac{\sqrt{13}-7}{18}\right)\left(x-\frac{-\sqrt{13}-7}{18}\right)$$

Baholash

$$9x^{2}+7x+1$$