x^2+10x+5 eching

$$x ^ { 2 } + 10 x + 5$$

Omil

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

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.

$$x^{2}+10x+5=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5}}{2}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 5}}{2}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-20}}{2}$$

100 ni -20 ga qo'shish.

$$x=\frac{-10±\sqrt{80}}{2}$$

80 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±4\sqrt{5}}{2}$$

x=\frac{-10±4\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -10 ni 4\sqrt{5} ga qo'shish.

$$x=\frac{4\sqrt{5}-10}{2}$$

-10+4\sqrt{5} ni 2 ga bo'lish.

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

x=\frac{-10±4\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -10 dan 4\sqrt{5} ni ayirish.

$$x=\frac{-4\sqrt{5}-10}{2}$$

-10-4\sqrt{5} ni 2 ga bo'lish.

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

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

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

Baholash

$$x^{2}+10x+5$$