x^2+-10x+20 eching

$$x ^ { 2 } + - 10 x + 20$$

Omil

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

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.

$$x^{2}-10x+20=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 20}}{2}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 20}}{2}$$

-4 ni 20 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-80}}{2}$$

100 ni -80 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{20}}{2}$$

20 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{5}}{2}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{5}}{2}$$

x=\frac{10±2\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{5} ga qo'shish.

$$x=\frac{2\sqrt{5}+10}{2}$$

10+2\sqrt{5} ni 2 ga bo'lish.

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

x=\frac{10±2\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{5} ni ayirish.

$$x=\frac{10-2\sqrt{5}}{2}$$

10-2\sqrt{5} ni 2 ga bo'lish.

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

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

$$x^{2}-10x+20=\left(x-\left(\sqrt{5}+5\right)\right)\left(x-\left(5-\sqrt{5}\right)\right)$$

Baholash

$$x^{2}-10x+20$$