x^2-2x-4 eching

$${ x }^{ 2 } -2x-4$$

Omil

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

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.

$$x^{2}-2x-4=0$$

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.

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

-2 kvadratini chiqarish.

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

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+16}}{2}$$

4 ni 16 ga qo'shish.

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

20 ning kvadrat ildizini chiqarish.

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

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{5}}{2}$$

x=\frac{2±2\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{5} ga qo'shish.

$$x=\frac{2\sqrt{5}+2}{2}$$

2+2\sqrt{5} ni 2 ga bo'lish.

$$x=\sqrt{5}+1$$

x=\frac{2±2\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{5} ni ayirish.

$$x=\frac{2-2\sqrt{5}}{2}$$

2-2\sqrt{5} ni 2 ga bo'lish.

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

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

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

Baholash

$$x^{2}-2x-4$$