10x^2+25x+5 eching

$$10 x ^ { 2 } + 25 x + 5$$

Omil

$$10\left(x-\frac{-\sqrt{17}-5}{4}\right)\left(x-\frac{\sqrt{17}-5}{4}\right)$$

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.

$$10x^{2}+25x+5=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 10\times 5}}{2\times 10}$$

25 kvadratini chiqarish.

$$x=\frac{-25±\sqrt{625-4\times 10\times 5}}{2\times 10}$$

-4 ni 10 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625-40\times 5}}{2\times 10}$$

-40 ni 5 marotabaga ko'paytirish.

$$x=\frac{-25±\sqrt{625-200}}{2\times 10}$$

625 ni -200 ga qo'shish.

$$x=\frac{-25±\sqrt{425}}{2\times 10}$$

425 ning kvadrat ildizini chiqarish.

$$x=\frac{-25±5\sqrt{17}}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$x=\frac{-25±5\sqrt{17}}{20}$$

x=\frac{-25±5\sqrt{17}}{20} tenglamasini yeching, bunda ± musbat. -25 ni 5\sqrt{17} ga qo'shish.

$$x=\frac{5\sqrt{17}-25}{20}$$

-25+5\sqrt{17} ni 20 ga bo'lish.

$$x=\frac{\sqrt{17}-5}{4}$$

x=\frac{-25±5\sqrt{17}}{20} tenglamasini yeching, bunda ± manfiy. -25 dan 5\sqrt{17} ni ayirish.

$$x=\frac{-5\sqrt{17}-25}{20}$$

-25-5\sqrt{17} ni 20 ga bo'lish.

$$x=\frac{-\sqrt{17}-5}{4}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-5+\sqrt{17}}{4} ga va x_{2} uchun \frac{-5-\sqrt{17}}{4} ga bo‘ling.

$$10x^{2}+25x+5=10\left(x-\frac{\sqrt{17}-5}{4}\right)\left(x-\frac{-\sqrt{17}-5}{4}\right)$$

Baholash

$$10x^{2}+25x+5$$