5x^2-32x+2 eching

$$5 { x }^{ 2 } -32x+2$$

Omil

$$5\left(x-\frac{16-\sqrt{246}}{5}\right)\left(x-\frac{\sqrt{246}+16}{5}\right)$$

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.

$$5x^{2}-32x+2=0$$

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.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 5\times 2}}{2\times 5}$$

-32 kvadratini chiqarish.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 5\times 2}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-32\right)±\sqrt{1024-20\times 2}}{2\times 5}$$

-20 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-32\right)±\sqrt{1024-40}}{2\times 5}$$

1024 ni -40 ga qo'shish.

$$x=\frac{-\left(-32\right)±\sqrt{984}}{2\times 5}$$

984 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-32\right)±2\sqrt{246}}{2\times 5}$$

-32 ning teskarisi 32 ga teng.

$$x=\frac{32±2\sqrt{246}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{32±2\sqrt{246}}{10}$$

x=\frac{32±2\sqrt{246}}{10} tenglamasini yeching, bunda ± musbat. 32 ni 2\sqrt{246} ga qo'shish.

$$x=\frac{2\sqrt{246}+32}{10}$$

32+2\sqrt{246} ni 10 ga bo'lish.

$$x=\frac{\sqrt{246}+16}{5}$$

x=\frac{32±2\sqrt{246}}{10} tenglamasini yeching, bunda ± manfiy. 32 dan 2\sqrt{246} ni ayirish.

$$x=\frac{32-2\sqrt{246}}{10}$$

32-2\sqrt{246} ni 10 ga bo'lish.

$$x=\frac{16-\sqrt{246}}{5}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{16+\sqrt{246}}{5} ga va x_{2} uchun \frac{16-\sqrt{246}}{5} ga bo‘ling.

$$5x^{2}-32x+2=5\left(x-\frac{\sqrt{246}+16}{5}\right)\left(x-\frac{16-\sqrt{246}}{5}\right)$$

Baholash

$$5x^{2}-32x+2$$