3x^2+5x+1? eching

$$3 x ^ { 2 } + 5 x + 1 ?$$

Omil

$$3\left(x-\frac{-\sqrt{13}-5}{6}\right)\left(x-\frac{\sqrt{13}-5}{6}\right)$$

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.

$$3x^{2}+5x+1=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3}}{2\times 3}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 3}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-12}}{2\times 3}$$

25 ni -12 ga qo'shish.

$$x=\frac{-5±\sqrt{13}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{13}}{6}$$

x=\frac{-5±\sqrt{13}}{6} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{13} ga qo'shish.

$$x=\frac{\sqrt{13}-5}{6}$$

x=\frac{-5±\sqrt{13}}{6} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{13} ni ayirish.

$$x=\frac{-\sqrt{13}-5}{6}$$

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{13}}{6} ga va x_{2} uchun \frac{-5-\sqrt{13}}{6} ga bo‘ling.

$$3x^{2}+5x+1=3\left(x-\frac{\sqrt{13}-5}{6}\right)\left(x-\frac{-\sqrt{13}-5}{6}\right)$$

Baholash

$$3x^{2}+5x+1$$