5x^2+2x-1 eching

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

Omil

$$5\left(x-\frac{-\sqrt{6}-1}{5}\right)\left(x-\frac{\sqrt{6}-1}{5}\right)$$

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.

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

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\times 5\left(-1\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4-20\left(-1\right)}}{2\times 5}$$

-20 ni -1 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+20}}{2\times 5}$$

4 ni 20 ga qo'shish.

$$x=\frac{-2±\sqrt{24}}{2\times 5}$$

24 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{6}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-2±2\sqrt{6}}{10}$$

x=\frac{-2±2\sqrt{6}}{10} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{6} ga qo'shish.

$$x=\frac{2\sqrt{6}-2}{10}$$

-2+2\sqrt{6} ni 10 ga bo'lish.

$$x=\frac{\sqrt{6}-1}{5}$$

x=\frac{-2±2\sqrt{6}}{10} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{6} ni ayirish.

$$x=\frac{-2\sqrt{6}-2}{10}$$

-2-2\sqrt{6} ni 10 ga bo'lish.

$$x=\frac{-\sqrt{6}-1}{5}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-1+\sqrt{6}}{5} ga va x_{2} uchun \frac{-1-\sqrt{6}}{5} ga bo‘ling.

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

Baholash

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