(2x^2+7x+4) eching

$$( 2 x ^ { 2 } + 7 x + 4 )$$

Omil

$$2\left(x-\frac{-\sqrt{17}-7}{4}\right)\left(x-\frac{\sqrt{17}-7}{4}\right)$$

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.

$$2x^{2}+7x+4=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 4}}{2\times 2}$$

7 kvadratini chiqarish.

$$x=\frac{-7±\sqrt{49-4\times 2\times 4}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49-8\times 4}}{2\times 2}$$

-8 ni 4 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49-32}}{2\times 2}$$

49 ni -32 ga qo'shish.

$$x=\frac{-7±\sqrt{17}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{17}}{4}$$

x=\frac{-7±\sqrt{17}}{4} tenglamasini yeching, bunda ± musbat. -7 ni \sqrt{17} ga qo'shish.

$$x=\frac{\sqrt{17}-7}{4}$$

x=\frac{-7±\sqrt{17}}{4} tenglamasini yeching, bunda ± manfiy. -7 dan \sqrt{17} ni ayirish.

$$x=\frac{-\sqrt{17}-7}{4}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-7+\sqrt{17}}{4} ga va x_{2} uchun \frac{-7-\sqrt{17}}{4} ga bo‘ling.

$$2x^{2}+7x+4=2\left(x-\frac{\sqrt{17}-7}{4}\right)\left(x-\frac{-\sqrt{17}-7}{4}\right)$$

Baholash

$$2x^{2}+7x+4$$