2x^2-9x+5 eching

$$2 x ^ { 2 } - 9 x + 5$$

Omil

$$2\left(x-\frac{9-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+9}{4}\right)$$

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.

$$2x^{2}-9x+5=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 5}}{2\times 2}$$

-9 kvadratini chiqarish.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 5}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 5}}{2\times 2}$$

-8 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81-40}}{2\times 2}$$

81 ni -40 ga qo'shish.

$$x=\frac{-\left(-9\right)±\sqrt{41}}{2\times 2}$$

-9 ning teskarisi 9 ga teng.

$$x=\frac{9±\sqrt{41}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{9±\sqrt{41}}{4}$$

x=\frac{9±\sqrt{41}}{4} tenglamasini yeching, bunda ± musbat. 9 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}+9}{4}$$

x=\frac{9±\sqrt{41}}{4} tenglamasini yeching, bunda ± manfiy. 9 dan \sqrt{41} ni ayirish.

$$x=\frac{9-\sqrt{41}}{4}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{9+\sqrt{41}}{4} ga va x_{2} uchun \frac{9-\sqrt{41}}{4} ga bo‘ling.

$$2x^{2}-9x+5=2\left(x-\frac{\sqrt{41}+9}{4}\right)\left(x-\frac{9-\sqrt{41}}{4}\right)$$

Baholash

$$2x^{2}-9x+5$$