2x^2+3x+17x+2 eching

$$2 x ^ { 2 } + 3 x + 17 x + 2$$

Omil

$$2\left(x-\left(-2\sqrt{6}-5\right)\right)\left(x-\left(2\sqrt{6}-5\right)\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

20x ni olish uchun 3x va 17x ni birlashtirish.

$$factor(2x^{2}+20x+2)$$

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.

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

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 2\times 2}}{2\times 2}$$

20 kvadratini chiqarish.

$$x=\frac{-20±\sqrt{400-4\times 2\times 2}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-20±\sqrt{400-8\times 2}}{2\times 2}$$

-8 ni 2 marotabaga ko'paytirish.

$$x=\frac{-20±\sqrt{400-16}}{2\times 2}$$

400 ni -16 ga qo'shish.

$$x=\frac{-20±\sqrt{384}}{2\times 2}$$

384 ning kvadrat ildizini chiqarish.

$$x=\frac{-20±8\sqrt{6}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-20±8\sqrt{6}}{4}$$

x=\frac{-20±8\sqrt{6}}{4} tenglamasini yeching, bunda ± musbat. -20 ni 8\sqrt{6} ga qo'shish.

$$x=\frac{8\sqrt{6}-20}{4}$$

-20+8\sqrt{6} ni 4 ga bo'lish.

$$x=2\sqrt{6}-5$$

x=\frac{-20±8\sqrt{6}}{4} tenglamasini yeching, bunda ± manfiy. -20 dan 8\sqrt{6} ni ayirish.

$$x=\frac{-8\sqrt{6}-20}{4}$$

-20-8\sqrt{6} ni 4 ga bo'lish.

$$x=-2\sqrt{6}-5$$

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{6} ga va x_{2} uchun -5-2\sqrt{6} ga bo‘ling.

$$2x^{2}+20x+2=2\left(x-\left(2\sqrt{6}-5\right)\right)\left(x-\left(-2\sqrt{6}-5\right)\right)$$

Baholash

$$2\left(x^{2}+10x+1\right)$$

Yechim qadamlari

20x ni olish uchun 3x va 17x ni birlashtirish.

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